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3 years ago

Write the net ionic equation for the reaction that occurs, if any, when gold is placed in a solution of hydrobromic acid

Chemistry

1 answer:

Gre4nikov [31]3 years ago

3 0

Answer:

ionic salt MX2

Explanation:

Atoms from an element in group IIA are allowed to react with atoms from an element in group VIIA. What type of compound is likely to form

elements from IIA are alkali earth metals (M) and need to lose

2 e^1- to become stable as M^2+

elements from VIIA ar halogens (X) and need to gain 1 e^1-

to become stable so one M^2+ combines with 2 X^1- to make

an ionic salt MX2

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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago

What is the primary requirement for a molecule to be analyzed by Gas Chromatography?

BabaBlast [244]

Answer:

b) The molecule has a molecular weight under 200 g/mole

Explanation:

The molecule has a molecular weight under 200 g/mole is the primary requirement for a molecule to be analyzed by Gas Chromatography.

3 0

3 years ago

Read 2 more answers

When a plant cell with 36 chromosomes goes through mitosis, how many daughter cells will be produced and how many chromosomes wi

SCORPION-xisa [38]

Answer:

I believe it would be 30 chromosomes

Explanation:

Hope that this helps you ;-)

7 0

3 years ago

Read 2 more answers

Which is the MOST precise measurement this cylinder gives for the volume of liquid it contains?

disa [49]

Hello! Before I answer to your question, please be sure to include a picture or else a report will be filed for your question would be commenced. You are lucky I have this for my homework tonight and I figured it out. Thank you:

The answer to your question would be as followed:

The most precise measurement for the cylinder you are mentioning is B. 43.0mL

6 0

3 years ago

Can anyone solve this please. I need this fastly

Natalka [10]

Answer:

Average atomic mass = 63.3896

Explanation:

Step 1: Find how much Cu-65 we have

1 - Amount of Cu-63 = Amount of Cu-65

1 - 0.6915 = Amount of Cu-65

Amount of Cu-65 = 0.305

Step 2: Find the average atomic mass of Cu

(0.6915 x 63) + (0.305 x 65) = Average atomic mass

(43.5645) + (19.825) = Average atomic mass

63.3895 = Average atomic mass

Therefore the average atomic mass of Cu is 63.3895 atomic mass units

4 0

3 years ago