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Otrada [13]
4 years ago
13

A cup of gold colored metal bed was measured to have a mask for 25 g. By water displacement, the volume of the bed was calculate

d to be 40.0 mL. Given the following densities, identify the metal. Gold equals 19.3 g/milliliters copper equals 8.86 g/milliliters bronze equals 9.87 g/milliliters

Chemistry
1 answer:
N76 [4]4 years ago
5 0

Answer : The metal is copper.

Explanation :

As we are given that:

Mass of metal = 25 g

Volume of metal = 40.0 mL

Formula used:

\text{Density of metal}=\frac{\text{Mass of metal}}{\text{Volume of metal}}

Now putting all the given values in this formula, we get:

\text{Density of metal}=\frac{25g}{40.0mL}

\text{Density of metal}=8.86g/mL

From this we conclude that the metal is copper whose density is 8.86 g/mL.

Hence, the metal is copper.

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Indicates the writer's stance on the main idea of a paragraph. The controlling idea appears in the paragraph's topic sentence. ... The final sentence of a paragraph that summarizes the topic sentence using different words. Words and phrases that show how the ideas in sentences and paragraphs are related.

Explanation:

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Hydrogen ions,H+ ions??
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A 1.50 g sample of solid NH₄NO₃ was added to 35.0 mL of water in a styrofoam cup (insulated from the environment) and stirred un
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Answer : The heat of the reaction is, 1.27 kJ/mole

Explanation :

First we have to calculate the heat released.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of sample = 1.50 g

c = specific heat of water = 4.81J/g^oC

T_1 = initial temperature  = 22.7^oC

T_2 = final temperature  = 19.4^oC

Now put all the given value in the above formula, we get:

Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC

Q=-23.8095J=-0.0238kJ

Now we have to calculate the heat of the reaction in kJ/mol.

\Delta H=\frac{Q}{n}

where,

\Delta H = enthalpy change = ?

Q = heat released = 0.0238 kJ

n = number of moles NH₄NO₃ = \frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=\frac{1.50g}{80g/mol}=0.01875mole

\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole

Therefore, the heat of the reaction is, 1.27 kJ/mole

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