Answer:
A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
Explanation:
Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.
Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C
However, when ethylene glycol is mixed with water in a ratio of 50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.
This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation:
Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.
Answer:
Q = 246 kJ
Explanation:
It is given that,
Mass of water, m = 200 g
Let initial temperature, [tex]T_i=5^{\circ}[/tex]
Final temperature of water, [tex]T_f=300^{\circ} C[/tex]
We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]
So, the heat energy needed to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]
or
Q = 246 kJ
So, the heat energy of 246 kJ is needed.
All the following are oxidation–reduction reactions except:________
a. H2(g) + F2(g) → 2HF(g).
b. Ca(s) + H2(g) → CaH2(s).
c. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).
d. 6Li(s) + N2(g) → 2Li3N(s).
e. Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).
Answer:
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)
Explanation:
All the following are oxidation–reduction reactions except:________
a. H₂(g) + F₂(g) → 2HF(g). Redox. H is oxidized and F is reduced.
b. Ca(s) + H₂(g) → CaH₂(s). Redox. Ca is oxidized and H is reduced.
c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Redox. K is oxidized and H is reduced.
d. 6Li(s) + N₂(g) → 2Li₃N(s). Redox. Li is oxidized and N is reduced.
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number
The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.
Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.
The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).
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A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that
[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]
where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula
[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]
After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was present 8.008.00 days ago
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g
Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.
Answer:
Here's what I get
Explanation:
pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44
The higher the pKₐ, the weaker the acid.
Thus, o-vanillin is the weaker acid and has a stronger conjugate base.
The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.
The equation for the equilibrium is
H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺
weaker acid weaker base stronger base stronger acid
The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.
Thus, o-vanillin does not fully protonate p-toluidine.
O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.
The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.
The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.
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Which of these are elimination reactions? Check all that apply.
CH3OH + CH3COOH → CH3CO2CH3 + H20
C3H7OH → C3H6 + H20
H9C2Br + NaOH → C2H4 + NaBr + H20
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
B and C. Just finished my lesson on Edge.
Predict the product of the following Wittig reaction. Be sure your answer accounts for stereochemistry, where appropriate. If multiple stereoisomers form, be sure to draw all products using appropriate wedges and dashes.
1. PPh3
5-iodo-1-phenyl-1-pentanone →
2. n- BuLi
Answer:
Final product: cyclopent-1-en-1-ylbenzene
Explanation:
In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).
See figure 1
I hope it helps!
Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below
Answer:
afshkkyfugutuiryfyi
If the SN2 reaction of an aromatic alcohol with an alkyl halide, like the synthesis of nerolin, is successful, what changes would be seen in the IR spectrum for the product compared to the starting material
Answer:
O-H stretch signal at 3300 cm-1
Explanation:
In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).
In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).
I hope it helps!
Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?
Write a balanced chemical equation to support your answer.
Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
assume that amonia can be prepared by the folowing reaction in the gas phase at STP. If the reaction conditions are maintainted at STP, how many liters of NH3 can be produced by the reaction of 12.0 L of H2 and the exact required volumen of N2
Answer:
8.00L of ammonia can be produced
Explanation:
The reaction is:
N₂(g) + 3H₂(g) → 2NH₃(g)
Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.
As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:
1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia
Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:
12L H₂(g) ₓ (2L NH₃(g) / 3L H₂(g)) =
8.00L of ammonia can be producedWhat is the empirical formula for the compound: C8H8S2?
Answer:
Empirical formula = C4H4SExplanation:
The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.
The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____
Answer:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
The reactant that is reduced is Q.
Explanation:
The complete equation for the reaction is such that:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently, [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].
From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.
Calculate the mass percent of .485g of H, which reacts with O to form 2.32g H2O?
Answer:
53.1% of hydrogen reacts
Explanation:
The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.
The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:
Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water
1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:
0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H
In mass:
0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water
As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:
(Mass that reacts / Mass added) × 100
(0.260g / 0.485g) × 100 =
53.1% of hydrogen reactsoxygen get stable configuration by ____________two electrons
please give the answer as fast as you can
please
Answer:
gaining two electrons
Explanation:
electron configuration
2:6
so add two to 6 to get stable 2:8
question attached ! asap please
Answer:
A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.
b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules
c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.
d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.
e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids
Explanation:
True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.
Answer:
Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.
Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added
Answer:
The correct answer is 1.60.
Explanation:
Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,
Moles = volume * concentration of HCl
= 25/1000*0.10 = 0.0025 moles
Similarly the moles of NaOH added will be determined by using the formula,
Moles of NaOH added = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
= 0.001 / 0.040 = 0.025 M
pH = -log[H+]
= -log[0.025]
= 1.60
The pH after 15 ml of NaOH has been volume is 1.60.
Calculation of Concentration of HCl Moles
It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,
Moles is = volume * concentration of HCl
Then is = 25/1000*0.10 = 0.0025 moles
Besides, the moles of NaOH added will be determined by using the formula,
When the Moles of NaOH added is = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
When The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
After that = 0.001 / 0.040 = 0.025 M
pH = -log[H+]
Then = -log[0.025]
Therefore, = 1.60
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1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
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HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
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Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)
Answer:
B
Explanation:
The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.
Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Round your answer to significant digit.
Answer:
Molarity Cu²⁺ = 0.423M Cu²⁺
Explanation:
40.8g of copper (II) acetate into 200mL of a 0.700M sodium chromate
The reaction of copper acetate with sodium chromate occurs as follows:
Cu(CH₃COO)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2CH₃COONa
In water, the Copper(II) acetate dissociates in Cu²⁺ cation.
To know final molarity of Cu²⁺ we need to calculate the moles of Cu²⁺ that don't react with chromate ion, thus:
Moles of 40.8g of copper(II) acetate (Molar mass: 181.63g/mol) are:
40.8g × (1mol / 181.63g) = 0.2246 moles of Copper(II) acetate
Moles of sodium chromate are:
0.200L ₓ (0.700mol / L) = 0.140 moles of sodium chromate.
As 1 mole of Copper(II) acetate reacts per mole of sodium chromate, moles of Copper(II) acetate = Moles of Cu²⁺ that remains after the reaction are:
0.2246mol - 0.140moles = 0.0846 moles of Cu²⁺
Molarity is ratio between moles of solute (Moles Cu²⁺) and volume in liters of solution (200mL = 0.200L):
Molarity Cu²⁺ = 0.0846 moles / 0.200L
Molarity Cu²⁺ = 0.423M Cu²⁺Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Answer:
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Explanation:
Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature, K = Constant
You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:
[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]
In this case:
P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 KReplacing:
[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]
Solving:
[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]
P2=1.44 atm
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Answer all to the best of your ability please
Answer:
2–Ethyl–3–methlypentanal.
Explanation:
To name the compound given in the question above, we must observe the following:
1. The functional group of the compound is Alkanal i.e Aldehyde,
—CHO and it is located at carbon 1.
Note: the aldehyde functional group is always at carbon 1 and there will be no need to state it's position in the compound.
2. The longest continuous carbon chain is 5 i.e pentane. But the presence of the functional group will replace the –e at the end of pentane with –al, making the name to the pentanal.
3. The substituents attached are:
a. Ethyl, CH2CH3 at carbon 2.
b. Methyl, CH3 at carbon 3.
4. Combine the above to get the name of the compound.
Therefore, the name of the compound is:
2–Ethyl–3–methlypentanal.
For the reaction system, 2SO2(g) + O2(g) <--> 2SO3(g), the equilibrium concentrations are: SO3: 0.120M SO2: 0.860M O2: 0.330M Calculate the value of Kc for this reaction.
Answer:
0.0590 M⁻¹
Explanation:
Kc represents the equilibrium constant. It is given as;
Kc = [products] / [reactants]
For the reaction; 2SO2(g) + O2(g) <--> 2SO3
Products = SO3
Reactants = SO2 and O2
Kc is given as;
Kc = [SO3]² / [SO2]² [O2]
Kc = 0.120² / (0.860)² (0.330)
Kc = 0.0144 / 0.2440 = 0.0590 M⁻¹
Question 14 of 25
What type of reaction is BaCl2 + Na,504 → 2NaCl + Baso,?
A. Single-replacement
B. Synthesis
C. Double-replacement
D. Decomposition
double displacement
bcoz each of the reactants combines with other reactants to obtain the product
Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 CReplacing:
Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C
Solving:
Q=3,294.9 J
The heat absorbed by the sample of water is 3,294.9 J