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ehidna [41]
3 years ago
13

4. Nitrogen and oxygen gases react to form dinitrogen monoxide gas (N2O). What volume of O, is

Chemistry
1 answer:
solmaris [256]3 years ago
8 0

Answer:

no be t ha t s had ice

Explanation:

uah an

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Ralph's friend invited him to attend a hard rock concert. Ralph did not want to go because he assumed other people who attended
Iteru [2.4K]

Answer:

Prejudice

Explanation:

8 0
3 years ago
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Assuming all other variables are constant, if the pressure of a gas is increased from 2. atm to 6 atm the volume would change fr
Leona [35]

Answer:

2.0 Liters

Explanation:

Boyles law

P1 V1 = P2 V2

V2= P1 V1 / V2

7 0
3 years ago
What is the bond angle in a tetrahedral molecule?
levacccp [35]
109.5

tetrahedral shape:
number of electron pair = 4,
number of bonded pair = 4,
number of lone pair = 0.
7 0
3 years ago
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How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?
LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

0.500 L \times \frac{0.360mol}{L} = 0.180 mol

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

0.180 mol \times \frac{110.98g}{mol} = 20.0 g

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

4 0
2 years ago
A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are
ohaa [14]

<u>Answer:</u> The final volume of the oxygen gas is 4.04 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K

Putting values in above equation, we get:

\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L

Hence, the final volume of the oxygen gas is 4.04 L

8 0
4 years ago
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