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3 years ago

What is the molarity of solution that contains 5.80 grams of nickel nitrate,ni(no3)2 in 500.ml?

1 answer:

3 0

Molarity is defined as the number of moles of solute in 1 L of solution
molar mass of Ni(NO₃)₂ - 182.7 g/mol
number of moles of Ni(NO₃)₂ - 5.80 g/ 182.7 g/mol = 0.0317 mol
number of moles in 500 mL - 0.0317 mol
therefore number of moles in 1000 mL - 0.0317 mol / 500 mL x 1000 mL/L
molarity of the solution is - 0.0634 M

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2 years ago

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2 years ago

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Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

Hence, the correct answer is C.

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2 years ago