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Maslowich
3 years ago
5

What substance is given off when salt is heated gently?​

Chemistry
1 answer:
KiRa [710]3 years ago
8 0

Answer:

Water

Explanation:

When a salt solution is heated, the solvent (assumedly water) will evaporate at a greater rate because the water molecules have greater energy.

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If 1.2 moles of ideal gas occupy a volume of 18.2 L at a pressure of 1.8 ATM what is the temperature of gas in degrees Celsius.
Novosadov [1.4K]

Answer: A 59.5 degree celcius

The equation that we will use to solve this problem is :

PV = nRT where:

P is the pressure of gas = 1.8 atm

V is the volume of gas = 18.2 liters

n is the number of moles of gas = 1.2 moles

R is the gas constant =  0.0821

T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:

(1.8)(18.2) = (1.2)(0.0821)(T)

T = 332.5 degree kelvin

The last step is to convert the degree kelvin into degree celcius:

T = 332.5 - 273 = 59.5 degree celcius

8 0
4 years ago
(a) Calculate the mass of CaCl2·6H2O needed to prepare 0.125 m CaCl2(aq) by using 500. g of water. (b) What mass of NiSO4·6H2O m
Alinara [238K]

Answer:

(a) 13.7 g.

(b) 28.91 g.

Explanation:

  • molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.

∴ m = (no. of moles of solute)/(mass of water (kg))

<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>

<em />

<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>

∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).

m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).

∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.

<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>

∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).

m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).

∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.

6 0
4 years ago
How do you do part B?
galina1969 [7]
I cannot see the whole equation.Therefore I shall not answer

7 0
3 years ago
Predict what might happen if the human body did not have specialized cells, tissues, organs, and organ systems to maintain homeo
bekas [8.4K]
We would be dead. No heart, brain or blood flow <span />
3 0
3 years ago
Read 2 more answers
A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

Hence, the volume of concentrated solution required is 9.95 mL

6 0
3 years ago
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