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2 years ago

Determine the mass in grams of each element.

Chemistry

1 answer:

jeka57 [31]2 years ago

7 0

1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g

2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g

3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g

4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g

<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>

Mole of Sb = 1.33×10²² mole

Molar mass of Sb = 122 g/mol

Mass of Sb =?

Mass = mole × molar mass

Mass of Sb = 1.33×10²² × 122

Mass of Sb = 1.62×10²⁴ g

<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>

Mole of Pt = 4.75×10¹⁴ mole

Molar mass of Pt = 122 g/mol

Mass of Pt =?

Mass = mole × molar mass

Mass of Pt = 4.75×10¹⁴ × 195

Mass of Pt = 9.26×10¹⁶ g

<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>

Mole of Ag = 1.22×10²³ mole

Molar mass of Ag = 108 g/mol

Mass of Ag =?

Mass = mole × molar mass

Mass of Ag = 1.22×10²³ × 108

Mass of Ag = 1.32×10²⁵ g

<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>

Mole of Cr = 9.85×10²⁴ mole

Molar mass of Cr = 52 g/mol

Mass of Cr =?

Mass = mole × molar mass

Mass of Cr = 9.85×10²⁴ × 52

Mass of Cr = 5.12×10²⁶ g

Learn more about mole:

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3 years ago

Read 2 more answers

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$