Answer: The answer is Strontium(Sr) i.e there are 38 electrons in this electronic configuration.
How To Find: You should count the number of electrons in each orbitals.
For example:
1s2 has two electrons in its orbital.Count similarly the number of electrons for every orbital.
This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas
Answer:
Maximum mass of compound produced = 64 Kg
Explanation:
Carbon monoxide react with hydrogen to produce methanol. Equation for the reaction is given as follows:
CO(g) + 2H₂(g) -----> CH₃OH(l)
From the equation of reaction, 1 mole of CO reacts with 2 moles of Hydrogen gas to produce 1 mole of methanol.
Molar mass of CO = 28 g
Molar mass of H₂ = 2 g
Molar mass of methanol = 32 g
Therefore, 28 g of CO reacts with 4 g of H₂ to produce 32 g of methanol
8.0 Kg of hydrogen will react with 8 * (28/4) Kg of CO = 56 Kg of CO.
therefore, hydrogen is the limiting reactant in the reaction under consideration.
8.0 Kg of Hydrogen will react with 56 Kg of CO to produce 8 * (32/4) Kg of methanol = 64 Kg of methanol
Therefore, maximum mass of compound produced = 64 Kg
Gallium chloride is formed by the reaction of 2.25 L of a 1.50 M solution of HCl according to the following equation: 2Ga 6HCl --> 2GaCl3 3H2 Determine the mass of gallium chloride, in grams, produced. Group of answer choices
Answer:
198.56g of GaCl3
Explanation:
We'll begin by calculating the number of mole HCl in 2.25 L of a 1.50 M solution of HCl. This is illustrated below:
Molarity of HCl = 1.50 M
Volume = 2.25 L
Mole of HCl =..?
Molarity = mole /Volume
1.5 = mole /2.25
Cross multiply
Mole = 1.5 x 2.25
Mole of HCl = 3.375 mole
Next, we shall determine the number of mole Gallium chloride, GaCl3 produced from the reaction. This is shown below:
2Ga + 6HCl —> 2GaCl3 + 3H2
From the balanced equation above,
6 moles of HCl reacted to produce 2 moles of GaCl3.
Therefore, 3.375 mole of HCl will react to produce = (3.375 x 2)/6 = 1.125 mole of GaCl3.
Therefore, 1.125 moles of GaCl3 were produced from the reaction.
Next, we shall convert 1.125 mole of GaCl3 to grams. This is illustrated below:
Molar mass of GaCl3 = 70 + (35.5x3) = 176.5g/mol
Mole of GaCl3 = 1.125 mole
Mass of GaCl3 =..?
Mole = mass /Molar mass
1.125 = mass of GaCl3 /176.5
Cross multiply
Mass of GaCl3 = 1.125 x 176.5
Mass of GaCl3 = 198.56g
Therefore, 198.56g of GaCl3 were produced from the reaction.
Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the solution?
Answer:
1.811 g
Explanation:
The computation of the mass need to use to make the solution is shown below:
We know that molarity is
[tex]Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}[/tex]
So,
[tex]Number\ of\ moles = Molarity\ \times Volume\ in\ L[/tex]
[tex]= 0.223\times 0.141[/tex]
= 0.031 moles
Now
[tex]Mass = moles \times Molecualr\ weight[/tex]
where,
The Molecular weight of NaCl is 58.44 g/mole
And, the moles are 0.031 moles
So, the mass of NaCL is
[tex]= 0.031 \times 58.44[/tex]
= 1.811 g
We simply applied the above formulas
You need to purify 2.0 grams of an impure sample of Acetanilide. The sample is contaminated with aniline. After the purification is complete you isolate 0.8 grams of acetanilide and record a melting point range of 108-110 °C. Complete the following calculations and show your work.
a. Calculate the minimum amount of distilled water you would use to complete the recrystallization.
b. How much acetanilide will still be dissolved in solution even after the sample is cooled to 0 °C?
c. Calculate the % recovery and the % error for the melting point.
d. Why is the percent recovery less than 100%? Describe multiple sources for loss of sample.
Answer:
Following are the answer to this question:
Explanation:
In the given question an attachment file is missing, that is attached. please find the attached file, and the following are the description of the given points:
a. At 100 degrees in 100 mL 5 g is dissolved.
For, it required:
[tex]\to 2g = 100 \times \frac{2}{5}[/tex]
[tex]= 40 \ \ ml \ of \ water.[/tex]
b. At 0 degrees 100 mL dissolve in 0.3 g.
So, the dissolve:
[tex]\to 40 \ ml= 0.3\times \frac{40}{100}[/tex]
[tex]= 0.12g.[/tex]
After refrigeration 0.12 g will still be dissolved.
c. After dissolving and freezing, precipitation can occur which would still be impure if the cooling is instantaneous. The added solvent was also too hard to recrystallize. The solvent was placed below its place of reservation.
d. Recovery percentage:
[tex]\to \frac{0.8}{2}\times100[/tex]
[tex]\to 40 \ \%[/tex]
The melting point of acetanilide:
[tex]\to 114.3^{\circ}.[/tex]
Found=109(medium)
Melting point error percentages:
[tex]= \frac{114.3-109}{114.3}\\\\=4.63 \ \%[/tex]
A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.0 femtoliters. This particular type of bacteria is known to communicate with its own species by secreting a small molecule called bactoX ( MW=126.9 g/mol ). A. Each bacterium contains 7140 bactoX molecules that can be secreted. How many moles of bactoX are present in a 3.0 μL sample volume that contains 7.512×106 bacterial cells?
Answer:
There are [tex]\mathbf{8.90172 \times 10^{-14}}[/tex] moles of bactoX present in a 3.0 μL sample volume that contains 7.512×106 bacterial cells
Explanation:
Given that:
The number of molecules present in one bacterial cell is [tex]7.140 \times 10^3[/tex] molecules
and the sample contains [tex]7.512 \times 10^6[/tex] molecules.
Number of moles = number of molecules /Avogadro's number
where;
Avogadro's number = 6.023 × 10²³
Number of moles = [tex]\dfrac{7.140 \times 10^3}{6.023 \times 10^{23}}[/tex]
Number of moles = [tex]1.185 \times 10^{-20}[/tex] moles
So; [tex]1.185 \times 10^{-20}[/tex] moles is present in one bacteria cell
Similarly; the sample contains [tex]7.512 \times 10^6[/tex] molecules.
Therefore; the number of moles present in the bactoX is = [tex]1.185 \times 10^{-20} \times 7.512 \times 10^6[/tex]
= [tex]\mathbf{8.90172 \times 10^{-14}}[/tex] moles
what is the best course of action if solid material remains in the flask after the heating step of recrystallization
Answer:
filter the hot mixture.
Explanation:
Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling
The process by which the movement of internal bonds converts one type of what organic compound into another is
Answer: It's is called Reaarrangement
Explanation:
Rearrangement is a chemic reaction in which an atom or group of atoms bond migrate from one carbon atom to another. The movement involves two adjacent atoms and move over a large distance. It is important in functional group transformation.
Reaarrangement can be classified based on the atom or group of atoms it migrate.
Nucleophilic rearrangement which involves the migration of atoms with electron pairs.
Electrophilic migration involved the migration of atom or groups without electron pair.
Free radical involves the atom or group of atoms moving with a lone pair of electron.
Calculate the mass percent by volume of 330.1 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) in 325 mL of solution.
Answer: The mass percent by volume is 101.6%
Explanation:
The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.
It is represented in formula as :
mass percent by volume =[tex]\frac{\text {mass of solute}\times 100}{\text {Volume of solution in ml}}\%[/tex]
Given : mass of glucose = 330.1 g
volume of solution = 325 ml
Thus mass percent by volume =[tex]\frac{330.1g\times 100}{325ml}=101.6\%[/tex]
Thus the mass percent by volume is 101.6%
What is the mass number of an atom with 24 protons and 30 neutrons?
Answer:
54
Explanation:
Mass number = protones + neutrons
Mass number = 24 + 30
Mass number = 54
PLEASE ANSWER AS SOON AS POSSIBLE REALLY WOULD APPRECIATE IT
Answer:
The answer is option D.
Hope this helps you
what is the reduction half equation of Fe(s)+ 2 HC2H3O2(aq) → Fe(C2H3O2)2(aq) + H2(g)
Answer:
2 H⁺ + 2e = H₂ ( reduction )
Explanation:
Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂
Fe( s ) = Fe⁺² + 2e ( oxidation )
2 H⁺ + 2e = H₂ ( reduction )
A sample of gas in a cylinder as in the example in Part A has an initial volume of 48.0 L , and you have determined that it contains 1.80 moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 12.0 L . How many moles of gas (n2) remain in the cylinder
Answer:
0.45 moles
Explanation:
The computation of the number of moles left in the cylinder is shown below:
As we know that
[tex]\frac{n1}{V1} = \frac{n2}{V2}[/tex]
we can say that
[tex]n2 = n1 \times \frac{V2}{V1}[/tex]
where,
n1 = 1.80 moles of gas
V2 = 12.0 L
And, the V1 = 48.0 L
Now placing these values to the above formula
So, the moles of gas in n2 left is
[tex]= 1.80 \times \frac{12.0\ L}{48.0\ L}[/tex]
= 0.45 moles
We simply applied the above formulas so that the n2 moles of gas could arrive
Carbon dioxide gas reacts with liquid water to produce aqueous carbonic acid.” Which chemical equation correctly translates this description? CO2 (s) + H2O (g) → H2CO3 (s) CO2 (l) + H2O (l) → H2CO3 (l) CO2 (g) + H2O (g) → H2CO3 (aq) CO2 (g) + H2O (l) → H2CO3 (aq)
Answer: CO2(g)+H2O(I) > H2CO3(Aq)
Explanation:
Got it right?
What is the maximum number of electrons in the second principal energy level?
02 32 8 18
Answer:
8 electrons
Explanation:
The second principal energy level has two sublevels: 2s and 2p
2s : 2 electrons
2p : 6 electrons (3 sublevels × 2 electrons each = 6 electrons)
It can hold a maximum of 8 electrons.
Hope this helps. :)
Convert 120 degrees F to K.
[?]K
Answer:
322
Explanation:
This is easy
Write a balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product.
Answer:
2 CH₃OH + 3 O₂ ⇒ 2 CO₂ + 4 H₂O
Explanation:
Methanol is CH₃OH. Oxygen is O₂. A combustion produces CO₂ and H₂O. Create an equation using this information and balance.
CH₃OH + O₂ ⇒ CO₂ + H₂O
2 CH₃OH + 3 O₂ ⇒ 2 CO₂ + 4 H₂O
The balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product is
CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)
From the question,
We are to write a balanced equation for the combustion of liquid methanol in air.
The combustion of liquid methanol in air is the reaction between methanol (CH₃OH) and oxygen (O₂). The reaction yields carbon(IV) oxide and water.
Now, for the balanced equation for the combustion of liquid methanol in air
The balanced chemical equation is
CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)
Hence, the balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product is CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)
Learn more here: https://brainly.com/question/2473060
How much MnO2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 715 Torr g
THIS IS THE COMPLETE QUESTION
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation.
How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL of Cl2(g) at 25 °C and 715 Torr?
Answer:
0.62901mol of MnO2(s) should be added
Explanation:
Given:
P = 715/760 = 0.94078atm
v=195ml=0.195l
n = ? moles have to find
R = 0.0821 L atm/K/mole
T = 25 + 273 = 298 K
Then we will make use of below formula
PV = nRT
Insert the values
0.94078*0.195=n 0.0821*298
24.466n=0.1740443
n=0.174/24.466
n=0.007235 nb of moles of cl2
as 1 mole of Cl2 were obtained from 1 mole of MnO2
so 0.007235 of chlorine must have come from
0.007235 moles of MnO2
1 mole of MnO2 = 86.94 g/mole
so 0.007235 moles of MnO2== 86.94* 0.007235
=0.62901
Which of the following is an inorganic compound?
Na4C
C2H6
C12H22O11
CH3COOH
Answer:
The answer is option A.
Na4C
Hope this helps you
4Ga + 3S2 ⇒ 2Ga2S3
How many grams of Gallium Sulfide would form if 20.5 moles of Gallium burned?
Answer:
2415.9g (corrected to 1 d.p.)
Explanation:
(Take the atomic mass of Ga=69.7 and S=32.1)
Assuming Ga is the limiting reagent (because the question did not mention the amount of sulphur burnt),
From the balanced equation, the mole ratio of Ga:Ga2S3 = 4: 2 = 2: 1, which means, every 2 moles of Ga burnt, 1 mole of Ga2S3 is produced.
Using this ratio, let y be the no. of moles of Ga2S3 produced,
[tex]\frac{2}{1} =\frac{20.5}{y}[/tex]
y = 20.5 / 2
= 10.25 mol
Since mass = no. of moles x molar mass,
the mass of Ga2S3 produced = 10.25 x (69.7x2 + 32.1x3)
= 2415.9g (corrected to 1 d.p.)
NH4NO2(s)→N2(g)+H2O(l) ---------------- Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
The balanced equation is :
NH4NO2(solid) = N2(gas) + 2 H2O(liquid)
Explanation:
A balanced chemical equation is an equation that has an equal number of atoms and charges on both sides of the equation. The given equation in question is imbalanced as the number of atoms not equal.
In this reaction, solid ammonium nitrite breaks into nitrogen gas and water, reaction known as decomposition.
The correct and balanced equation as follows :
NH4NO2(s) = N2(g) + 2 H2O(l)
Five mol of calcium carbide are combined with 10 mol of water in a closed, rigid, high-pressure vessel of 1800 cm3 internal empty volume. Acetylene gas is produced by the reaction:
Answer:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂
Explanation:
In order to find out the reaction, we must know the reactants.
For this situation, we make acetylene gas from carbide calcium CaC₂ and H₂O (water); therefore the reactants are:
- CaC₂ and H₂O
Acetylene is one of the products made → C₂H₂
So the reaction can be formed as this: CaC₂ + H₂O → C₂H₂
We missed the calcium, and this reaction also makes, Calcium Hydroxide, so the complete equation must be:
CaC₂ + H₂O → C₂H₂ + Ca(OH)₂
This is unbalanced, because we have 1 O in left side and 2 in right side so we add 2 in water so now, we get the complete reaction:
1 mol of calcium carbide reacts to 2 mol of water in order to produce 1 mol of acetylene and 1 mol of calcium hydroxide.
PLEASE HELP, 5 STARS, 35 POINTS + PRIZE The reaction C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol. At 600.0 K, the rate constant, , is 6.1×10−8 s−1. What is the value of the rate constant at 780.0 K?
Answer:
346 hope this is helpfull
What is a heterogeneous mixture?
Answer:
The type of mixture whose components are seen through our naked eyes is known as heterogeneous mixture. it is a mixture of small constituent parts of substances.
for eg, mixture of sand and sugar.
hope it helps..
Calculate the mass in grams for 0.251 moles of Na2CO3
Answer:
Explanation:
the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole
106*0.251=26.606 grames
Calculate the volume of 0.500 M C2H3O2H and 0.500 M C2H3O2Na required to prepare 0.100 L of pH 5.00 buffer with a buffer strength of 0.100 M. The pKa of C2H3O2H is 4.75.
Answer:
You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂H
Explanation:
It is possible to obtain pH of a weak acid using H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
For the buffer of acetic acid/acetate, the equation is:
pH = pKa + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]
Replacing:
5.00 = 4.75 + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]
1.7783 = [C₂H₃O₂Na] / [C₂H₃O₂H] (1)
Buffer strength is the concentration of the buffer, that means:
0.1M = [C₂H₃O₂Na] + [C₂H₃O₂H] (2)
Replacing (2) in (1):
1.7783 = 0.1M - [C₂H₃O₂H] / [C₂H₃O₂H]
1.7783 [C₂H₃O₂H] = 0.1M - [C₂H₃O₂H]
2.7783 [C₂H₃O₂H] = 0.1M
[C₂H₃O₂H] = 0.036MAlso:
[C₂H₃O₂Na] = 0.1M - 0.036M
[C₂H₃O₂Na] = 0.064MThe moles of both compounds you require is:
[C₂H₃O₂Na] = 0.1L × (0.064mol / L) = 0.0064moles
[C₂H₃O₂H] = 0.1L × (0.036mol / L) = 0.0036moles
Your stock solutions are 0.500M, thus, volume of both solutions you require is:
[C₂H₃O₂Na] = 0.0064moles × (1L / 0.500M) = 0.0128L = 12.8mL
[C₂H₃O₂H] = 0.0036moles × (1L / 0.500M) = 0.0072mL = 7.2mL
You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂HFor each of the following ground state electron configurations, determine what is incorrect. N: 1s²2s¹2p³ A) The configuration is not in the ground state. B) The configuration is missing an electron in the 2s orbital. C) The configuration has too many electrons in the 2p orbital.
Answer:
B
Explanation:
It should be 2s² which means the answer is the configuration is missing an electron in the 2s orbital.
Answer:
B.
Explanation:
It should be 2s² which means the answer
is the configuration is missing an electron
in the 2s orbital.
Lead can be prepared from galena [lead(II) sulfide] by first heating with oxygen to form lead(II) oxide and sulfur dioxide. Heating the metal oxide with more galena forms the metal and more sulfur dioxide. Write a balanced equation for the overall reaction by adding the balanced equations for the two steps.
Answer:
2 PbS(s) + 1.5 O₂(g) + PbO(s) ⇒ 2 SO₂(g) + 3 Pb(s)
Explanation:
Lead can be prepared from galena [lead(II) sulfide] by first heating with oxygen to form lead(II) oxide and sulfur dioxide. The corresponding chemical equation is:
PbS(s) + 1.5 O₂(g) ⇒ PbO(s) + SO₂(g)
Heating the metal oxide with more galena forms the metal and more sulfur dioxide. The corresponding chemical equation is:
2 PbO(s) + PbS(s) ⇒ 3 Pb(s) + SO₂(g)
We can get the overall reaction by adding both steps and canceling what is repeated on both sides.
2 PbS(s) + 1.5 O₂(g) + 2 PbO(s) ⇒ PbO(s) + 2 SO₂(g) + 3 Pb(s)
2 PbS(s) + 1.5 O₂(g) + PbO(s) ⇒ 2 SO₂(g) + 3 Pb(s)
4. What are the potential sources of error that might cause disagreement between the activity series' prediction of reactions and your observations of reactions
Answer:
1. Not to have enough salt water on the foil
2.not cleaning the foil well to remove interfering materials
Using the volumes of EDTA solution you just entered and the corresponding dry unknown sample masses entered earlier, calculate the percent mass of calcium carbonate in the unknown sample mixture.
Enter the calculated percent mass of calcium carbonate in the dry unknown sample for each of the 3 acceptable trials.
Be sure to enter your mass percentages to the correct number of significant digits and in the corresponding order that you entered your masses of your dry unknown samples and volumes of your EDTA previously. The dry unknown sample mass you entered for entry #1 below should correspond to the percent mass of calcium carbonate you enter for entry #1 here.
Trial #: Mass (Grams):
#1: 0.015
#2: 0.015
#3: 0.015
Volume (mL)
#1: 16.4
#2: 15.00
#3: 18.70
Molarity of EDTA Solution: 0.0675
Answer:
#1
Explanation:
molarity of EDTA solution 0.0675
no1
A sample of gas occupies a volume of 7.50 L at 0.988 atm and 301 K. At what temperature is the volume of the gas 4.00 L if the pressure is kept constant.
Answer:
160.53L
Explanation:
Since Pressure is kept constant we can use charles law
V1/T1 =V2/T2