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3 years ago

HELP ASAP

Chemistry

2 answers:

noname [10]3 years ago

5 0

Answer:

3.01 L

Explanation:

: 2.75L

∘

: ?

∘

If you know your gas laws, you have to utilize a certain gas law called Charles' Law:

is the initial volume,

is initial temperature,

is final volume,

is final temperature.

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

2.75

291

318

By cross multiplying, we have...

291

= 874.5

Dividing both sides by 291 to isolate

, we get...

= 3.005...

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

Therefore,

= 3.01 L

Anestetic [448]3 years ago

4 0

V2=3.01 L
explanation

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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

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Explanation:

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For Sample 1:

Total mass of sample = 100 g

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Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

For Sample 2:

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

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