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4 years ago

What are the coefficients that belong in the blanks to balance the following chemical equation?

Chemistry

1 answer:

brilliants [131]4 years ago

4 0

Answer:

2AgI + 1Na₂S ⟶ 1Ag₂S +2NaI

Explanation:

silver(I) iodide + sodium sulfide ⟶ silver(I) sulfide + sodium iodide

1. Convert the word equation to a chemical equation:

AgI + Na₂S ⟶ Ag₂S + NaI

2. Put a 1 in front of the most complicated-looking formula (Na₂S?):

AgI + 1Na₂S ⟶ Ag₂S + NaI

3. Balance Na:

We have fixed 2 Na on the left. We need 2 Na on the right. Put a 2 in front of NaI.

AgI + 1Na₂S ⟶ Ag₂S +2NaI

4. Balance S:

We have fixed 1 S on the left. We need 1 O on the right. Put a 1 in front of Ag₂S.

AgI + 1Na₂S ⟶ 1Ag₂S +2NaI

5. Balance Ag:

We have fixed 2Ag on the right. We need 2 Ag on the left. Put a 2 in front of AgI.

2AgI + 1Na₂S ⟶ 1Ag₂S +2NaI

Every formula now has a coefficient. The equation should be balanced.

6. Check that atoms balance:

Atom On the left On the right

Ag 2 2

I 2 2

Na 2 2

S 1 1

The balanced equation is

2AgI + 1Na₂S ⟶ 1Ag₂S +2NaI

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If a bag of peas weighs 454 grams, how many peas would be in the bag?

OverLord2011 [107]

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.1 g} = 4540pea$

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.36 g} \approx 1261pea$

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

You can learn more about conversion factors here: brainly.com/question/1844638

6 0

3 years ago

Determine the mass in grams of 125 mol of neon

Vesnalui [34]

Molar mass of Neon ( Ne ) = 20.1797 g/mol

m = n * mm

m = 125 * 20.1797

m = 2522.4625 g

hope this helps!

5 0

3 years ago

a 50.0 mL sample of KCl requires 22.40 mL of 0.0229 M Pb(NO3)2 in order to completely titrate it. What is the Molarity of the KC

garik1379 [7]

Answer:

0.02052M

Explanation:

First, we need to write a balanced equation for the reaction. This is illustrated below:

2KCl + Pb(NO3)2 → 2KNO3 + PbCl2

The following were obtained from the question:

Molarity of Pb(NO3)2 = 0.0229M

Volume of Pb(NO3)2 = 22.40 = 22.4/1000 = 0.0224L

Number of mole of Pb(NO3)2 =?

Recall:

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.0229x0.0224

Mole of Pb(NO3)2 = 5.13x10^-4mole

From the equation,

1mole of Pb(NO3)2 required 2moles KCl.

Therefore, 5.13x10^-4mole of Pb(NO3)2 will require = 5.13x10^-4x2 = 1.026x10^-3mole of KCl.

Now we can use this amount (i.e 1.026x10^-3mole) to find the molarity of KCl. This is illustrated below:

Mole of KCl = 1.026x10^-3mole

Volume of KCl = 500mL = 50/1000 = 0.05L

Molarity =?

Molarity = mole /Volume

Molarity of KCl = 1.026x10^-3/0.05

Molarity of KCl = 0.02052M

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4 years ago

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uysha [10]

Answer:

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4 years ago

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miss Akunina [59]

Answer:

yes

Explanation:

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