Find the missing part of this equation

How many atoms of zinc are in 6.05 moles of zinc

stiv31 [10]

As we know that one mole contains 6.022 × 10²³ particles. And the number of particles are related to moles as,

# of Particles = Moles × 6.022 × 10²³ particles

In case of Zinc atoms,

# of Atoms = Moles × 6.022 × 10²³ Atoms/mol

Putting value of moles.

# of Atoms = 6.05 mol × 6.022 × 10²³ Atoms/mol

# of Atoms = 3.64 × 10²⁴ Atoms

4 0

3 years ago

There are many careers in the area of

Lina20 [59]

Number 2. Tectonophysicist

3 0

4 years ago

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}$

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

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j

nydimaria [60]

Answer:

Cp = 0.237 J.g⁻¹.°C⁻¹

Explanation:

Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 640 J

m = mass = 125 g

Cp = Specific Heat Capacity = <u>??</u>

ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 640 J / (125 g × 21.6 °C)

Cp = 0.237 J.g⁻¹.°C⁻¹

3 0

3 years ago

Why do scientists use models? a. Models make some things easier to understand. b.Measurements can be made only on models. c. Mod

mina [271]

In science, a model is a representation of an idea, an object or even a process or a system that is used to describe and explain phenomena that cannot be experienced directly. Models are central to what scientists do, both in their research as well as when communicating their explanations.

So your answer would be

C) Models help scientists visualize things they can’t see.

8 0

3 years ago

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