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ladessa [460]
1 year ago
5

Which of the following below is the complete ionic equation of the reaction: Calcium nitrate and sodium sulfide solutions react

to form solid calcium sulfide and sodium nitrate solution

Chemistry
1 answer:
3241004551 [841]1 year ago
5 0

To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.

The balanced equation of the reaction will be:

Ca(NO_3)_{2(aq)}+Na_2S_{(aq)}\rightarrow CaS_{(s)}+2NaNO_{3(aq)}

Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +

Ca_{(aq)}^{2+}+2(NO_3)_{(aq)}^-+2Na_{(aq)}^++S_{(aq)}^{2-}\operatorname{\rightarrow}CaS_{(s)}+2Na_{(aq)}^++2NO^-_{3(aq)}

So, the answer will be option A

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What is the volume of 0.80 grams of O2 gas at STP? (5 points) Group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
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Answer:

0.56 liters

Explanation:

First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:

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At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:

  • 0.025 mol * 22.4 L/mol = 0.56 L

Thus the answer is 0.56 liters.

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in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

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Answer:Hello How was your day been today :)

Explanation:

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Read 2 more answers
How would i balance this equation?
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Our coefficients would be:
C12H22O11: 1
O2: 12
CO2: 12
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Why?
With the coefficients, our reactants would be:
12 carbons, 22 hydrogens, and 35 oxygens.

To obey the law of conservation of mass, we would have to have an equal amount of each of those elements on the products side, since so much of that element needs to form an equal amount on the other side.

So, with those coefficiens, our products would be:
12 carbons total, 35 oxygens total, and 22 hydrogens total. If we compare to the reactants side, they are all equal!

And, since 1 or 11 cannot be simplifyed any smaller, we know the equation is completely balanced.

Hope I could help!
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