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3 years ago

Volume of 100ml. it’s density is 0.15g/ml. what is its mass?

Chemistry

2 answers:

11Alexandr11 [23.1K]3 years ago

3 0

$V=100mL\\
d=0,15\frac{g}{mL}\\\\
m=dV=0,15\frac{g}{mL}*100mL=15g$

Veseljchak [2.6K]3 years ago

3 0

Answer:

The mass of the substance is 15 grams.

Explanation:

Mass of the substance = m

Volume of the substance = V = 100 mL

Density of the substance = d = 0.15 g/mL

The formula of density is ;

$Density=\frac{Mass}{Volume}$

$d=\frac{m}{V}$

$0.15 g/mL=\frac{m}{100 mL}$

$m=0.15 g/mL\times 100 mL=15 g$

The mass of the substance is 15 grams.

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A sample of Mo(NO3)6 has 2.22 x 10^22 nitrogen atom, how many oxygen atoms does the sample have?

goldfiish [28.3K]

There are 3.98 × 10^23 atoms of oxygen in the sample.

Given that;

1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen

x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen

x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms

x = 0.0368 moles

The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18

Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.

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1 year ago

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Characteristics of acid

-It tastes sour
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Characteristics of base

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2 years ago

Give two examples of diffusion process in plants and animals.

lisabon 2012 [21]

answer:

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Explanation:

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2 years ago

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Misha Larkins [42]

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

$moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

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3 years ago

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

erik [133]

Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

Finally, the mass percent is:

$\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%$

Best regards.

8 0

2 years ago