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rodikova [14]
3 years ago
10

The pressure in an engine cylinder increases from 7.6 atm to 13 atm as the gas is compressed from 0.50 L to 0.10 L. What is the

expected cylinder temperature if the gas was initially 15°C?
Chemistry
2 answers:
REY [17]3 years ago
8 0
T k = 15 + 273 = 288 K

4.6 / 13 => 0.353 atm

0.50 / 0.10 => 5 L

<span>(15 + 273) K x (13 atm / 7.6 atm) x (0.50 L / 0.10 L) 
</span>
<span>= </span>2463.15 K

<span>hope this helps!</span>
Mkey [24]3 years ago
8 0

Answer:

98.53K or 174.47°C

Explanation:

Initial Pressure, P1 = 7.6atm

Final Pressure, P2 = 13atm

Initial Volume, V1 = 0.50L

Final Volume, V2 = 0.10L

Initial Temperature, T1 = 15 +273 = 288K (Converting to kelvin temperature)

Final Temperature, T2 = ?

The equation relating all these parameters id the ideal gas equation.

This is given as;

(P1V1) / T1 = (P2V2) / T2

Upon solving for T2, we have;

T2 = P2V2T1 / PIV1

Inserting the values,

T2 = (13 * 0.10 * 288) / (7.6 * 0.5)

T2 = 374.4 / 3.8

T2 = 98.53K or 174.47°C

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7 0
2 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
How many moles are in 4.85 x 10^25 atoms of carbon?
Serjik [45]

Answer:

4.85 x 10^25

Explanation:

thats what i was told by a calculator

6 0
2 years ago
Select the correct answer.
Xelga [282]
Lead (II) chloride+ potassium nitrate
PbCI2+KNO3(B)
6 0
2 years ago
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