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iVinArrow [24]
3 years ago
5

What is the temperature, in degrees Celsius, of a substance with a temperature of 49K? –322°C –224°C 224°C 322°C

Chemistry
1 answer:
vichka [17]3 years ago
4 0

Answer:

-224ºC

Explanation:

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Neporo4naja [7]

Answer:

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3 years ago
How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 65.1 moles of water vapor?
ivolga24 [154]
 <span>CO2 + 4 H2 -----> CH4 + 2H2O ......

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5 0
3 years ago
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ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea
kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

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

H2(g) + I2(g) ⇌ 2HI(g)     ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

<u>Step 2</u>: Calculate ΔG

ΔG = ΔG° + RTln Q  

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q  

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)  

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0
4 years ago
Ionic bonds are strong; therefore, ionic compounds
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8 0
3 years ago
Read 2 more answers
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

8.625\times \frac{1}{2} =120.5

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

7 0
3 years ago
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