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3 years ago

What is the formula of tin (II) acetate?

Chemistry

2 answers:

mash [69]3 years ago

8 0

Answer:

Sn(C2H3O2)2

Explanation:

C2H3O2 is Acetate

Sn is tin

Acetate has a -1 charge

In order for tin to have a +2 charge, we need two acetates

So we write it as (C2H3O2)2

You just leave Sn blank because of the roman numeral and since it is a transition metal.

So the answer is Sn(C2H3O2)2

Hoochie [10]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf Sn (C_2 H_3 O_2)_2}}$

Explanation:

First, find what tin and acetate is.

Acetate: C₂H₃O₂
Tin: Sn

Next, find the charge on acetate, which is -1.

The Roman Numeral 2 in parentheses after tin (II), signifies that the tin has a charge of +2.

We need to balance the charges, so they will equal 0.

1 acetate= -1
Tin = +2

The best way to balance is by using 2 acetate molecules

2 acetate = -2
Tin = +2
-2 +2=0

If we want 2 acetate molecules, we must add a subscript of 2 after acetate.

Acetate: C₂H₃O₂

2 Acetate: (C₂H₃O₂)₂

Add the tin (Sn) in front

Tin (II) Acetate: Sn (C₂H₃O₂)₂

The formula for tin (II) acetate is Sn (C₂H₃O₂)₂

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kow [346]

Answer:

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Explanation:

Step 1: Data given

Number of moles CuO = 70.8 moles

Number of moles NH3 = 40.1 moles

Molar mass CuO = 79.545 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles

There will remain 70.8 - 60.15 = 10.65 moles CuO

Step 3: Calculate moles Cu

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

For 40.1 moles NH3 we'll have 60.15 moles Cu

The theoretical yield of Cu(s) in moles is 60.15 moles

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Hello!

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dezoksy [38]

Answer:

The answer is

Explanation:

In order to calculate the volume needed we use the formula

$V_2 = \frac{C_1V_1}{C_2}$

where

C1 is the concentration of the stock solution

V1 is the volume of the stock solution

C2 is the concentration of the diluted solution

V2 is the volume of thevdiluted solution

From the question

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So we have

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We have the final answer as

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