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Lynna [10]
3 years ago
5

What is the effect of the following on the volume of 1 mol of an ideal gas? The temperature changes from 305 K to 32°C and the p

ressure changes from 2 atm to 101 kPa.
A. Volume increases by 200%
B. Volume decreases to 50% of its original value
C. Volume decreases by 100%
D. Volume increases by 100%
E. Volume does not change
Chemistry
1 answer:
ryzh [129]3 years ago
7 0

Answer:

The volume increases by 100%.

Explanation:

<u>Step 1:</u> Data given

Number of moles ideal gas = 1 mol

Initial temperature = 305 K

Final temperature = 32°C + 273.15 = 305.15 K

Initial pressure = 2 atm

final pressure = 101 kPa = 0.996792 atm

R = gasconstant = doesn't change

V1 = initial volume

V2= the final volume

<u>Step 2: </u>Calculate volume of original gas

P*V = n*R*T

(P*V)/ T = constante

(P1 * V1) / T1 = (P2 * V2)/ T2

In this situation we have:

(2atm * V1)/ 305  = (0.996792 *V2) / 305.15

0.006557*V1 = 0.003266*V2

V2 = 2*V1

We see that the final volume is twice the initial volume. So the volume gets doubled. The volume increases by 100%.

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It's very simple.

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3 years ago
Connie was making sodium chloride by adding an acid to an alkali. She followed the progress of the reaction with a pH sensor. At
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⇒Answer:

When the pH sensor hits pH=7.

⇒Explanation:

Because pH=7  is the indicator that the acid and alkali have been neutralized.

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2 years ago
For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No
Pachacha [2.7K]

Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

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3 years ago
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Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
dmitriy555 [2]

<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aluminium oxide:</u>

Given mass of aluminium oxide = 12.5 kg = 12500 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

  • <u>For NaOH:</u>

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

  • <u>For HF:</u>

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol

For the given chemical reaction:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with (6\times 122.6)=735.6mol of sodium hydroxide and (12\times 122.6)=1471.2mol of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = \frac{2}{1}\times 122.6=245.2mol of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

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