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cricket20 [7]
3 years ago
8

The reaction takes place in water. What happens to the equilibrium when the pressure is increased? a)It favors formation of prod

ucts. b)It is conserved. c)It favors formation of reactants. d)It does not change
Chemistry
1 answer:
OLga [1]3 years ago
6 0

Answer:

I don't really get the options but it favoures the reactant side.

Explanation:

Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g

If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.

With the explanations I have made, I hope the question is now clear to you.

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If 35.0 mL of water in a graduated cylinder is displaced by 8.00 mL
sdas [7]

Answer:

<h2>0.52 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}\\

From the question

volume = final volume of water - initial volume of water

volume = 35 - 8 = 27 mL

We have

density =  \frac{14}{27}  \\  = 0.518518...

We have the final answer as

<h3>0.52 g/mL</h3>

Hope this helps you

3 0
2 years ago
Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


4 0
2 years ago
A gas has a density of 1.57 g/L at 40.0 °C and 2.00 atm of pressure. What is the identity of the gas?
Naddika [18.5K]

Answer:

Neon

Explanation:

Step 1: Given and required data

  • Density of the gas (ρ): 1.57 g/L
  • Temperature (T): 40.0°C
  • Pressure (P): 2.00 atm
  • Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

6 0
2 years ago
PLZ HELP I WILL GIVE U BRAINLEST PLZ HELP!!!!!!!!!!
sergeinik [125]
My answer is A. I'm probably wrong. in bad in this subject
7 0
2 years ago
Read 2 more answers
Calculate the number of chlorine atoms that are present in 66.05 g of dichloromethane, ch2cl2. when you have the number, take it
Whitepunk [10]
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of chlorine = 35.5 grams
Therefore,
molar mass of CH2Cl2 = 12 + 2(1) + 2(35.5) = 85 grams

number of moles = mass / molar mass
number of moles of CH2Cl2 = 66.05 / 85 = 0.777 moles

One mole of CH2Cl2 contains two moles of Cl and each chlorine mole has Avogadro's number of atoms in it.
Therefore,
number of chlorine atoms in 0.777 moles of CH2Cl2 can be calculated as follows:
number of atoms = 0.777 * 2 * 6.022 * 10^23 = 9.358 * 10^23 atoms

Now, we will take log base 10 for this number:
log (9.358 * 10^23) = 23.97119
5 0
2 years ago
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