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3 years ago

NH3

Chemistry

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

First one is: ammonia

Second one is: calcium hydroxide

Explanation:

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Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago

When 120 g of carbon reacts completely with 320 g of oxygen the mass of carbon dioxide formed will be?

olga2289 [7]

440 cause mass cant be created or destroyed

6 0

3 years ago

How do I do question number 3? And what is the answer?

shutvik [7]

Answer:

60.02 g.

Explanation:

It is clear from the balanced equation:

Mg + 2HCl → MgCl₂ + H₂.

that 1.0 mole of Mg reacts with 2.0 moles of HCl to produce 1.0 mole of MgCl₂ and 1.0 moles of H₂.

20.0 g of Mg reacts with excess HCl. To calculate the no. of grams of HCl that reacted, we should calculate the no. of moles of Mg:

no. of moles of Mg = mass/atomic mass = (20.0 g)/(24.3 g/mol) = 0.823 mol.

From the balanced equation; every 1.0 mol of Mg reacts with 2 moles of HCl.

∴ 0.833 mol of Mg will react with (2 x 0.833 mol = 1.646 mol) of HCl.

∴ The mass of reacted HCl = no. of moles x molar mass = (1.646 mol)(36.46 g/mol) = 60.02 g.

3 0

3 years ago

The reaction 2k(s) + br2(l) → 2kbr(s) is a(n) ______________ reaction.

Zinaida [17]

An oxidation reaction

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What do you think are some of the most pressing environmental science issues

Nezavi [6.7K]

Answer:

i believe global warming

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