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Hitman42 [59]
2 years ago
14

A gas has a volume of 5.00 L at 20 degrees Celsius. What is the volume of gas at 150 degrees Celsius.

Chemistry
1 answer:
WARRIOR [948]2 years ago
5 0

Answer:

7.22 L

Explanation:

From the question,

Applying Charles law,

V₁/T₁ = V₂/T₂...................... Equation 1

Where V₁ = Initial volume of gas, V₂ = Final volume of gas, T₁ = Initial Temperature of gas in Kelvin, T₂ = Final Temperature of gas in Kelvin

Make V₂ the subject of the equation

V₂ = (V₁×T₂)/T₁................. Equation 2

Given: V₁ = 5.00 L, T₁ = 20 °C = (20+273) K = 293 K, T₂ = 150 °C = (150+273) K = 423 K

Substitute these value into equation 2

V₂ = (5.00×423)/293

V₂ = 7.22 L

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The Mr is the mass numbers of each element added up so…. Fe = 56, O=16, H=1 … now add these up with the number of each element -> there’s 1 Fe, and 3 Os and 3 Hs as they are in brackets with a 3 outside-> (56+16+16+16+1+1+1=107) … your answer is 107
8 0
3 years ago
A sample of aluminum absorbs 9.86 J of heat, upon which the temperature increases from 23.2°C to 30.5°C. Since the specific heat
True [87]

Answer:

1.50 g

Explanation:

The heat absorbed by the aluminum in this case is:

q = m x C x ΔT     m= q/ (C x ΔT)

q= 9.86 J

C = 0.90 J/g-K

ΔT = ( 30.5 ºC - 23.2 ºC ) = 7.3 ºC = 7.3 K (this is a range of temperature)

m = 9.86 J / ( 0.90 J/g-K ) x 7.3 K ) = 1.50 g

8 0
3 years ago
How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at −57.1 °c. give your answer as a po
kolezko [41]
To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g; 
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C 
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348 
 = 26250.0348 J or 26.250 kJ
3 0
3 years ago
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
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