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3 years ago

Calculate the amount of heat needed to increase the temperature of 250 g of water from 20.0 ºC to 46.7 ºC.

1 answer:

8 0

The amount of heat needed to increase the temperature of 250 g water from 20 c to 46.7 c is calculated as follows

Q = MC delta T , where
Q= heat = ?
M = mass =250 g
C= specific heat capacity of water = 4.186 j/g c
delat T = change in temperature = 46.7 -20=26.7 c

Q = 250 g x4.186 j/g/c x26.7 c = 27941.55 j which is approximate to
28000 j(answer A)

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High concentrations of ammonia (NH3), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these specie

kykrilka [37]

Explanation:

It is known that molality is the number of moles present in kg of solution.

Mathematically, Molality = $\frac{\text{no. of moles of solute}}{\text{mass of solvent in Kg}}$

The given data is as follows.

Molar mass of ammonia = 17 g/mol

Concentration = 1.002 mg/L = $\frac{0.001002 g/L}{17 g/mol}$

= $5.89 \times 10^{-4}$ mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

Molality = $\frac{5.89 \times 10^{-4} mol/L}{1 kg/L}$

= $5.89 \times 10^{-4} mol/kg$

And,

Molar mass of nitrite = 46 g/mol

Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

= $8.956 \times 10^{-6}$ mol/L

And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

= $8.956 \times 10^{-6}$ mol/kg

Now, Molar mass of nitarte = 62 g/mol

Concentration = 1352.2 mg/L

= $\frac{1.3522 g/L}{62 g/mol}$

= 0.02181 mol/L

Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality will be calculated as follows.

Molality = $\frac{0.02181 mol/L}{1 kg/L}$

= 0.02181 mol/kg

Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$ for ammonia, $8.956 \times 10^{-6}$ mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.

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3 years ago

What is the density of ice at -10°C?

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Answer:

3.98 C my friend you welcome

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3 years ago

PLEASE HELP! Count the total number of atoms in H 2 O: 2 3 4 5 6

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3
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