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Korvikt [17]
3 years ago
15

How many calories are required to raise the temperature of 105 g of water from 30.0°c to 70.0°c?

Chemistry
1 answer:
ioda3 years ago
5 0
<span>4.200 calories. The formula is rise in heat (H) = the specific heat capacity (c) * rise in heat in degrees Fahrenheit (T) and volume of liquid (m) . H= cTm. Water, the typical comparison, has a specific heat capacity of 1. So to raise the Fahrenheit temperature of 105 grams of water 40 degrees, you multiply 105 * 40 * 1 = 4,200.</span>
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Black_prince [1.1K]
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3 years ago
1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?
RSB [31]

Answer : The number of molecules present in nitrogen gas are, 3.48\times 10^{13}

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of N_2 gas = 1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm      (1 atm = 760 mmHg)

V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 0.0^oC=273+0.0=273K

Now put all the given values in above equation, we get:

(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

Therefore, the number of molecules present in nitrogen gas are, 3.48\times 10^{13}

8 0
3 years ago
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lutik1710 [3]

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What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
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and when we have this equation:
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3 0
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