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3 years ago

By what process does water vapor become a cloud?

Chemistry

2 answers:

lions [1.4K]3 years ago

8 0

The answer is evaporation

Ivanshal [37]3 years ago

6 0

The answer is c. condensation

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A worker at a nuclear power plant has noticed that his dosimeter alarm has

Sphinxa [80]

Answer:

Radiation is being released from the reactor.

Explanation:

( A P E X )

5 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.

hoa [83]

The actual yield is 43 g Cl₂.

The limiting reactant was MnO₂ because it gave the smaller mass of Cl₂.

∴ The theoretical yield is 60.25 g Cl₂.

% yield = actual yield/theoretical yield × 100 %

Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g

8 0

3 years ago

Be sure to answer all parts. calculate δg ocell for the reaction between cr(s) and cu2+(aq). e ocell = 1.08 j/c. enter your answ

Mademuasel [1]

Answer:

$\boxed{-6.29 \times10^{5}\text{ J}}$

Explanation:

Step 1. Determine the cell potential

E°/V

2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V

3×[Cu²⁺ + 2e⁻ ⟶ Cu] 0.3419 V

2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V

Step 2. Calculate ΔG°

$\Delta G^{\circ} = -nFE_{\text{cell}}^{^{\circ}} = -6 \times 96 485 \times 1.086 = \text{-629 000 J}\\\\= \boxed{-6.29 \times10^{5}\text{ J}}$

6 0

3 years ago

What is 1520 9600 mg to kilograms

Ostrovityanka [42]

I Think that the answer is 15.2096 Kilograms, but I might be wrong.

5 0

3 years ago