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3 years ago

Which of the following experiments raises ethical concerns

Chemistry

1 answer:

xxMikexx [17]3 years ago

6 0

Answer:

Research that releases a poisonous gas into the air.

Explanation:

Since I don't know the options I will guess it is ^

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What is polar and unpolar bonds?

Mars2501 [29]

Answer:Nonpolar bonds form between two atoms that share their electrons equally. Polar bonds form when two bonded atoms share electrons unequally.

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3 years ago

How many moles of mgs2o3 are in 193 g of the compound?

Mashutka [201]

Answer:

1.414 Moles

Solution:

Data Given:

Mass of MgS₂O₃ = 193 g

M.Mass of MgS₂O₃ = 136.43 g.mol⁻¹

Moles = ?

Formula Used:

Moles = Mass ÷ M.Mass

Putting values,

Moles = 193 g ÷ 136.43 g.mol⁻¹

Moles = 1.414 mol

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3 years ago

Write a paragraph

steposvetlana [31]

I think It’s 55 but that’s just me

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3 years ago

What element(s) must be present in a molecule for it to be considered an organic molecule?

Burka [1]

Carbon is the one and only element to make a molecule organic.

4 0

3 years ago

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t

tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0

3 years ago