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Westkost [7]
3 years ago
5

What describes the behavior of an ideal gas?

Chemistry
1 answer:
valina [46]3 years ago
5 0

Answer:

The answer is D

Explanation:

for a p e x

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How many σ and π bonds are in this molecule? A chain of five carbon atoms. There is a double bond between the first and second c
kati45 [8]

Answer:

Sigma bonds: 10

Pi bonds: 4

Explanation:

The compound described must be CH2=CH-CO-CH≡CH. If we look at the compound closely, we will notice that there are 10 sigma bonds and 4 pi bonds.

There are three pi bonds between carbon atoms and one pi bond between a carbon and an oxygen atom (C=O). All these can easily be seen in the structure of the formula chosen in this answer.

5 0
4 years ago
An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/ml; = 62.07 g/mol) and water (d
tankabanditka [31]
a) Volume percent

Formula: % v/v = [volume solute / volume solution] * 100

Just to make it easy take a base of 50 volume parts of ethylen glycol and 50 volume parts of water to make 100 volumes of mixture (this assumpion will be valid for all the questions):

% v/v =[ 50 ml ethyleneglycol] / [100 ml mixture] * 100 = 50%

Answer: 50% v/v

b) Mass percent

% m/m = [mass ethylene glycol / mass solution] * 100

mass ethylene glycol = 50 ml * 1.114 g/ ml = 55.7 g

mass of mixture = 100 ml * 1.07 g/ml = 107 g

% m/m = [55.7 / 107 g] * 100 = 52.06 %

Answer: 52.06%

c) Molarity

M = number of moles of solute / liters of solution

number of moles of solute = mass in grams / molar mass

number of moles of ehtylene glycol = 55.7 g / 62.07 g/mol = 0.8974 mol

liters of solution = 0.1 liter

M = 0.8974 mol / 0.1 liter = 8.974 M

Answer: 8.974 M

d) Molality

m = number of moles of solute / kg of solvent

number of moles of ethylen glycol = 0.8974 mol

mass of water = 50 ml * 1 g/ml = 50 g = 0.05 kg

m = 0.8974 mol / 0.05 kg = 17.95 m

Answer: 17.95 m

e) mole fraction

mole fraction = [number of moles of solute] / [number of moles of mixture] * 100

number of moles of ethylen glycol = 0.8974 mol

number of moles of water = 50 g / 18.01 g /mol = 2.776 mol

mole fraction = 0.8974 mol / [0.8974 mol + 2.776 mol] = 0.244

Answer: 0.244
7 0
3 years ago
Where does energy go during endothermic and exothermic reactions?
ra1l [238]
It lecterns fries that is the answer.
7 0
3 years ago
What is the product of the unbalanced equation below?
inessss [21]

Answer:

C. HCI(g)

Explanation:

The following equation between hydrogen gas (H2) and oxygen gas (O2) is given below:

H2(g) + Cl2(g) ►

Based on these unbalanced equation, the products of the reaction was not given, however, if one molecule of hydrogen and oxygen combine, hydrogen chloride (HCl) should be produced as the product of the reaction as in:

H2(g) + Cl2(9) ► 2HCl(g)

4 0
3 years ago
When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

4 0
3 years ago
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