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djverab [1.8K]
3 years ago
14

How many calories are there in 32 Calories? a.64,000 b.16,000 C.32,000 D..032

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

32

Explanation:

There cannot be more in a number than the number. Therefore, the answer has to be D, or 32.

Hope this helps! :)

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The value of ΔG° at 181.0 °C for the formation of calcium chloride from calcium metal and chlorine gas is ________ kJ/mol. At 25
aleksandr82 [10.1K]

Answer:

\Delta G^{0} at 181.0 ^{0}\textrm{C} is -723.3 kJ/mol.

Explanation:

We know, \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}

where, T is temperature in kelvin.

Let's assume \Delta H^{0} and \Delta S^{0} does not change in the temperature range 25.0 ^{0}\textrm{C}  - 181.0 ^{0}\textrm{C}.

181.0^{0}\textrm{C} = (273+181.0) K = 454.0 K

Hence, at 181.0 ^{0}\textrm{C}, \Delta G^{0}=(-795.8kJ/mol)-[(454.0 K)\times (-159.8\times 10^{-3}kJ/K.mol)]

        = -723.3 kJ/mol

5 0
3 years ago
What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a
Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

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