Question 1 of 10
What do the conclusions tell about the experiment?
A. The conclusions tell how the experiment should be repeated.
B. The conclusions tell if the scientific method was followed.
C. The conclusions tell why the data support or reject the hypothesis.
D. The conclusions tell what other scientists think about the
experiment.

Answers

Answer 1

Answer:

C. The conclusions tell why the data support or reject the hypothesis.


Related Questions

A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calculation for the theoretical yield of Na2CO3.
What is the percent yield of sodium carbonate, Na2CO3?
6. A 1473-g unknown mixture with baking soda is heated and has a mass loss of 0.325 g. Refer to Example Exercise 14.2 and show the calculation for the percentage NaHCOs in the mixture.

Answers

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.

5. The theoretical yield of Na₂CO₃ is approximately 0.9809 gramsThe percentage yield of sodium carbonate is approximately 101.02%.

6. Percentage of NaHCO₃ in the mixture is approximately 59.76%.

Reasons:

Mass of baking soda = 1.555 g

Mass of Na₂CO₃ produced = 0.991 g

Required:

Calculation for the theoretical yield

Solution:

Theoretical yield (mass) of Na₂CO₃ produced is found as follows;

Molar mass of Na₂CO₃ = 105.9888 g/mol

Molar mass of NaHCO₃ = 84.007 g/mol

[tex]\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3[/tex]

The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.

The percentage yield is given as follows;

[tex]\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}[/tex]

The percentage yield of Na₂CO₃ is therefore;

[tex]\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}[/tex]

(Some baking soda may remain if the reaction is not completed)

6. Mass of the unknown mixture of baking soda = 1473 g

Mass loss from the mixture = 0.325 g

Required:

The percentage of NaHCO₃ in the mixture.

Solution:

The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃

Molar mass of H₂CO₃ = 62.03 g/mol

[tex]\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles[/tex]

The chemical reaction is presented as follows;

2NaHCO₃(s) [tex]\underrightarrow {\Delta \ Heated}[/tex] Na₂CO₃(s) + H₂CO₃(g)2 moles of NaHCO₃  produces 1 mole of H₂CO₃

The number of moles of NaHCO₃ in the mixture is therefore;

2 × 5.2394 × 10⁻³ moles ≈ 1.04788 × 10⁻² moles

Mass of NaHCO₃ in the mixture is therefore

Mass of NaHCO₃ = 1.04788 × 10⁻²  moles × 84.007 g/mol = 0.88029 g

[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}[/tex]

Which gives;

[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \ \frac{0.88029 \, g}{1.473 \, g} \times 100 \approx \underline{ 59.76 \%}[/tex]

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2. Using Tables 5 and 6 below, balance the following chemical reaction:
Na2CO3 (aq)+CaCl2 (aq) → CaCO3 (s)+NaCl (aq)

Answers

Answer:

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  2 NaCl

Explanation:

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  NaCl

First, determine what is balanced.  

Ca and CO₃ is balanced; there is one on each side.  

Na and Cl aren't balanced.  On one side, there is two of each, while on the other there is only one.  To fix this add a 2 in front of NaCl.  This will cause the equation to be balanced.

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  2 NaCl

Which of the following alkyl halides is most likely to undergo rearrangement in an SN1 reaction? Please explain your answer.
A) 3-­‐bromopentane
B) 2-­‐chloro-­‐3,3-­‐dimethylpentane
C) 3-­‐chloropentane
D) bromocyclohexane
E) 1-­‐bromo-­‐4-­‐methylcyclohexane

Answers

Ok okokokokoookokokokokokooko

lonic compounds can conduct electricity in

Answers

Answer:

Molten form

Explanation:

This is because they have mobile ions

How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?

Answers

Answer:

[tex]n_{Cl^-}=25molCl^-[/tex]

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

[tex]5\frac{molMgCl_2}{L}[/tex]

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

[tex]n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-[/tex]

Best regards.

A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.

Answers

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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A compound is found to contain 39.99 % carbon, 6.727 % hydrogen, and 53.28 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound

Answers

Answer:

CH₂O

Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of C:H:O.

Assume 100 g of the compound.

1. Calculate the mass of each element.

Then we have 39.99 g C, 6.727 g H, and 53.28 g O.

2. Calculate the moles of each element

[tex]\text{Moles of C} = \text{39.99 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.330 mol C}\\\\\text{Moles of H} = \text{6.727 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{6.674 mol H}\\\\\text{Moles of O} = \text{53.28 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{3.330 mol O}[/tex]

3. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

C:H:O = 3.330:6.674:3.330 = 1:2.004:1.000 ≈ 1:2:1

4. Write the empirical formula

EF = CH₂O

How are scientific questions answered?
A. Through observing and measuring the physical world
B. Through testing a theory about the physical world
c. Through forming a hypothesis about the question
D. Through predicting a solution about the question
SUBM

Answers

Answer:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

Answer: its A

through observing and measuring the physical world

Explanation:

Where would you find the following symbol on a diagram of a water molecule: δ+ ?

Answers

Answer:

Near the Hydrogen Atoms.

Explanation:

δ+ would be found near the dipole end of the hydrogens. δ- would be found near the dipole end of oxygen.

Why is not a good idea to drink seawater when people are lost at sea?
The semipermeable membrane protecting your stomach is ruptured during osmosis.
The osmotic pressure builds up in the cells of your intestine until they potentially rupture.
The high concentration of salt forces water out of the cells lining your stomach and intestine.
none of the above
oooo

Answers

The high concentration of salt forces water out of the cells lining your stomach and intestine.

Name of th molecule
1. CH3CH2CHClCHBrCH3
2.C=C-CH3
CH3CH=CHCH2

Answers

Answer:

1: 2-bromo-3-chloropentane

Explanation:

find longest carbon chain =5

place the Br and Cl on the carbon chain

follow naming rules I guess

Write the following isotope in nuclide notation: oxygen-14

Answers

Answer:

[tex]14\\8[/tex]O

Explanation:

The top number always represents the mass number.

The bottom number always represents the atomic number.

The element always goes after the numbers.

If charge is present, that comes after the element.

What did Rutherford discover that Thomson did not understand?

Answers

Answer: That most of an atom's mass was packed in a central nucleus

Answer:

That most of an atom's mass was packed in the central nucleus.

Explanation:

Rutherford disproved Thomson's model in 1911 with his gold foil experiment, in which he demonstrated that the atom has a high-mass nucleus.

How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water

Answers

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

In general, the solubility of a short chain carboxylic acid is greater than an ester with the same number of carbons. For example, butyric acid (four carbon atoms) is infinitely soluble in water, whereas ethyl acetate is only partially soluble in water (8 g/100 mL).

Select the statement that correctly explains why the water solubility of the carboxylic acid is greater than that of the ester.

a. Ethyl acetate has two -CH3 groups while butyric acid has only one.
b. The carboxylic acid is much more soluble because the-COOH group is on the end of the molecule and is therefore more accessible to H2O.
c. An ester group (-COO-) cannot form hydrogen bonds with water since there are no hydrogen atoms in the ester group.
d. Carboxylic acid groups (-COOH) can form hydrogen bonds with each other, but ester groups (Coo-) cannot.
e. A carboxylic ad roup CH) can form more hydrogen bonds with water than an ester group (-COO-).

Answers

Answer:

A carboxylic acid group (-COOH) can form more hydrogen bonds with water than an ester group (-COO-).

Explanation:

The carboxylic acid group (-COOH) is found in the carboxylic acids. This group is ultimately responsible for the solubility of carboxylic acids in water. It is worthy of note that the high boiling points of low molecular weight carboxylic acids is often because they are capable of intermolecular hydrogen bonding which leads to the dimerization of carboxylic acid.

The solubility of carboxylic acids decreases as the length of the alkyl chain increases. Hence, a long chain carboxylic acid is less soluble in water than shorter chain carboxylic acids.

Ester molecules can't form hydrogen bonds with each other but they do form weak hydrogen bonds with water. This leads to the solubility of low molecular weight esters. However, if a carboxylic acid and an ester posses the same length of alky chain, the carboxylic acid will form more hydrogen bonds and thus be more soluble in water than than a corresponding ester of the same chain length.

Question 23
1 pts
When solutions of AgNO3 and NaOH react, the balanced molecular equation is:
2 AgNO3(aq) + 2NaOH(aq) O--> Ag2O(s) + 2 NaNO3(aq) + H20(1)
How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?
a. 0.127 g
c. 0.273 g
b. 0.136 g
d. 0.400 g
OB
OC
OA
OD

Answers

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________

Answers

Answer:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

The solubility of O2 in water is approximately 0.00380 g L-1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760. torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?

Answers

Answer:

The correct answer is 0.00080 gram per liter.

Explanation:

Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.  

Based on Henry's law,  

Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas

Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas

= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr

= 0.00012 mol/L/atm

= 0.00012 M/atm

Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure

= 0.210 × 1 atm

= 0.210 atm

Now putting the values in Henry's law equation we get,  

Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,

= 0.000025 mol/L × 32 gram/mol

= 0.00080 gram per liter

Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:

Answers

The question is incomplete as some part is missing:

concentration in octanol Partition Ratio = concentration in water

a) What are the intermolecular forces of attraction between octanol molecules? Explain.

b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.

c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.

d) Would nonane (figure 2) be more soluble in water or octanol? Explain.

e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.

f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.

Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C

Answer:

1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.

2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).

3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.

4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.

5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.

The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.

Answers

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} }[/tex]

where,

R: ideal gas constantT: absolute temperatureM: molar mass of the gas

We can use the info of argon to calculate the temperature for both samples.

[tex]T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K[/tex]

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s[/tex]

Determine the rate of a reaction that follows the rate law:
rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1

Answers

Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

What is meant by rate of a reaction ?

Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.

Here,

The concentration of A, [A] = 1 M

The concentration of B, [B] = 3 M

The partial order with respect to A, m = 2

The partial order with respect to B, n = 1

The rate constant of the reaction, k = 1.5

The rate of the reaction,

r = k[A]^m [B}^n

r = 1.5 x 1² x 3

r = 4.5 mol L⁻¹s⁻¹

Hence,

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

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Determine the empirical formula of Magnesium Oxide from following data. Type the calculations.

Mass of Crucible and Cover + magnesium ribbon (before heating) =27.60 g

Mass of crucible and Cover = 27.30 g

Mass of magnesium metal =??

Mass of crucible and cover + magnesium oxide (after heating) = 27.80 g

Mass of combined oxide (after heating - before heating) =??

Answers

mass of magnesium metal = ( mass of crucible and cover + magnesium ribbon ( before heating) - (mass of crucible and cover )

mass of combined oxide = ( mass of crucible and cover + magnesium oxide) - (mass of crucible and cover + magnesium oxide) - (magnesium metal )

emprical formula :
find no of mole of Magnesium
= mass divide atomic number
= 0.0125

find for oxygen also
= 0.0125

ratio
0.0125 : 0.0125
1:1
MgO

The NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?

a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above

Answers

the answer to this problem is c
the answer is going to be “C. NaS (aq)” hope you have a good day and hope this helped

Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/

Answers

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

Which describes the molecule shown below?
A) Saturated Alkane
B) Saturated Alkene
C) Unsaturated Alkyne
D) Unsaturated Alkene
Please Help I'm stuck on this one!!!​

Answers

The molecule shown below is a Saturated Alkane, because are hydrocarbons that have only single bonds and open chains.

What are alkanes?

Alkanes are acyclic and saturated hydrocarbons, that is, they are compounds formed only by carbon and hydrogen atoms, open chain and with only single bonds between their carbons.

In this case, Saturated alkanes are hydrocarbons whose carbon atoms are linked only with single bonds.

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The molecule shown below is a Saturated Alkane, because are hydrocarbons that have only single bonds and open chains.

What are alkanes?

Alkanes are saturated hydrocarbons. This means that their carbon atoms are joined to each other by single bonds.

Alkanes are saturated hydrocarbons. This means that contains only carbon and hydrogen atoms bonded by single bonds only. The general formula for an alkane is [tex]C_nH_{2n+2}[/tex].

In this case, Saturated alkanes are hydrocarbons whose carbon atoms are linked only with single bonds.

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Give the concentration of each type of ion in the following solutions:
a. 0.50 M CO(NO3)2
b. 1 M Fe(C1O4)3​

Answers

Answer:

a.

[tex]M_{Co^{2+}}=0.5M\\ \\M_{NO_3^{-}}=1.0M[/tex]

b.

[tex]M_{Fe^{3+}}=1.0M\\ \\M_{ClO_4^{-}}=3.0M[/tex]

Explanation:

Hello,

a. In this case, the ions are cobalt (II) and nitrate, for which, one mole of cobalt (II) nitrate contains one mole of cobalt (II) and two moles of nitrate (see subscripts), therefore, concentrations turn out:

[tex]M_{Co^{2+}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{1molCo}{1molCo(NO_3)_2}=0.5M\\ \\M_{NO_3^{-}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{2molNO_3^{-}}{1molCo(NO_3)_2}=1.0M[/tex]

b. In this case, the ions are iron (III) and chlorate, for which one mole of iron (III) is contained in one mole of iron (III) chlorate and three moles of chlorate are in one mole of iron (III) chlorate (see subscripts), therefore, the concentrations turn out:

[tex]M_{Fe^{3+}}=1.0\frac{molFe(ClO_4)_3}{L}* \frac{1molFe^{3+}}{1molFe(ClO_4)_3}=1.0M\\ \\M_{ClO_4^{-}}=0.5\frac{molFe(ClO_4)_3}{L}* \frac{3molClO_4^{-}}{1molFe(ClO_4)_3}=3.0M[/tex]

Regards.

What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12

Answers

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

pH = 3.39

Consider the following precipitation reaction occurring in aqueous solution:
3 SrCl2(aq)+2 Li3PO4(aq) →Sr3(PO4)2(s)+6 LiCl(aq)

Write the complete ionic equation and the net ionic equation for this reaction.

Answers

Answer:

[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]

First equation is the complete ionic equation.

Second equation is the net ionic equation.

Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant

Answers

Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

What is a chemical equation?

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

Learn more about the chemical equation here:

https://brainly.com/question/12363199

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The next few questions will walk you through solving the following problem: At a given temperature, a 5.0M solution of hydrazine (N2H4) as a pH of 11.34. Hydrazine is base.
A. What is the concentration of hydroxide ion at equilibrium?
B. What is the pK for hydrazine reacting with water at this temperature?

Answers

Answer:

A. [OH⁻] = 2.188x10⁻³M

B. pKb = 6.02

Explanation:

When hydrazine is in equilbrium with water, its reaction is:

N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)

Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

A. From pH, you can find [OH⁻], thus:

pH = -log [H⁺]

11.34 = -log [H⁺]

4.57x10⁻¹² = [H⁺]

As 1x10⁻¹⁴ = [OH⁻] [H⁺]

1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]

[OH⁻] = 2.188x10⁻³M

B. Concentrations in equilibrium of the species are:

[N₂H₄] = 5.0M - X

[HN₂H₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate

As [OH⁻] = 2.188x10⁻³M

X = 2.188x10⁻³M

Replacing:

[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M

[HN₂H₄⁺] = 2.188x10⁻³M

[OH⁻] = 2.188x10⁻³M

Replacing in Kb expression:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]

Kb = 9.577x10⁻⁷

pKb is defined as -log Kb

pKb = -log 9.577x10⁻⁷

pKb = 6.02

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