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3 years ago

9

For each of the following compounds, write the symbols of the ions in the compound, and the number of each ion in the molecule o

f that compound.
Cal2
Na2CO3
Ga(CIO3)3
CuF2
(NH4)3PO4
FeSO4
Mg(NO3)2
NH4NO2
KC2H3O2
Na2Cr2O7

1 answer:

Kay [80]3 years ago

8 0

Ca^2+ and I^-
Na+ and Co3^2-
Ga^3+ and ClO3
Cu^2+ and F-
NH4^- and PO4^3-
Fe2+ and (SO4)^2-
Mg2+ and NO3^-
NH4^+ and NO2^-
K^+ and (C2H3O2)^- {C2H3O2 is acetate}
Na^+ and Cr2O7^2-

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GIVEN BRAINLIEST

umka21 [38]

Answer:

A. your observations in writing

Explanation:

Descriptive investigation is one of the mode of inquiries in science. It is devoid of the usual collection of data in the field or setting up laboratory experiments to compared variables.

In descriptive investigation, observations are usually done in writing.
They are usually common in Astrology.
Such investigation involves a question without a hypothesis ensuing.
Tests are usually not carried out in descriptive investigation.

6 0

3 years ago

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

3 years ago

What is the percent of hydrogen by mass in CH4O

WITCHER [35]

Answer:

My guess would be 12.5828

4 0

3 years ago

Rank these compounds by highest boiling point to lowest boiling point? Pentane, neopentane, isopentane

PilotLPTM [1.2K]

<span>Pentane has a boiling point of 36.1 C, while isopentane boils at 27.7C. Neopentane has the lowest boiling point at 9.5 C. Therefore from highest to lowest boiling points, it is pentane, isopentane, and neopentane.</span>

4 0

4 years ago

Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2

Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

Molar mass of AgNO3 = 169.87 g/mol
Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0

2 years ago