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3 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH 4NO 3 in order to prepare a 0.452 m solution?

Chemistry

1 answer:

Ivahew [28]3 years ago

6 0

Answer:

0.768 kg

Explanation:

Step 1: Given data

Mass of ammonium nitrate (solute): 27.8 g

Molality of the solution (m): 0.452 m

Step 2: Calculate the moles corresponding to 27.8 g of ammonium nitrate

The molar mass of ammonium nitrate is 80.04 g/mol.

$27.8 g \times \frac{1mol}{80.04g} = 0.347mol$

Step 3: Calculate the mass of water

The molality is equal to the moles of solute divided by the kilograms of solvent (water).

$m= \frac{n(solute)}{m(solvent)} \\m(solvent) = \frac{n(solute)}{m} = \frac{0.347mol}{0.452mol/kg} = 0.768 kg$

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Answer:

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Given data:

Number of moles of sugar = ?

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Solution:

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3 years ago

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

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$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

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3 years ago

Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

tia_tia [17]

Answer:

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Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

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