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3 years ago

Important (question on k12 test)

Chemistry

2 answers:

monitta3 years ago

4 0

Answer:

A new Dana system of classification contains 78 different classes of minerals based on composition and then further classified by type and group. To be considered a mineral, a substance must be an inorganic, naturally formed solid, with a specific chemical formula and a fixed internal structure. To test whether something is a mineral, there are several identification tests to which the substance is subjected, including its resistance to scratching, its density in comparison to water, its color, the degree of light it reflects, the color of the powdered mineral, its breakage pattern and its crystalline form.

Explanation:

Luden [163]3 years ago

4 0

Answer: chemical makeup and crystal lattice structure

Explanation:

The minerals are the inorganic substances that are extracted beneath the earth crust. These minerals are associated with the gangue particles that are required to be removed by processing.

The minerals can be classified on the basis of the chemical and physical properties. The chemical properties define the chemical composition of the mineral. The physical properties define how minerals react with the external environment.

Thus chemical makeup and crystal lattice structure can be a useful approach to classify the minerals as the minerals may differ chemically as well as in structural makeup.

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If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

2 years ago

What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead(ii) nitrate to precipitate all of the lead ions?

Svetradugi [14.3K]

Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.

5 0

3 years ago

The energy released by a nuclear fusion reaction is produced when

Igoryamba

Answer:

Explanation:

energy produced in nuclear fusion or even fission is due to mass defect and it can be calculated by Einstein equation e=mc2

8 0

3 years ago

Compute the freezing point of this solution:

likoan [24]

Answer:

Freezing point = 1.25

Explanation:

If we increase the concentration of the solution, the concentration of H+ does not change.

Convert 2.5% in to decimal

2.5% = 2.5 ÷100

= 0.025

The freezing point = 0.025 × 50

= 1.25

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3 years ago

What is the pH of a 0.40 M solution of C6H5COOH?

melomori [17]

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3 years ago