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VARVARA [1.3K]
4 years ago
11

What type of compound is ag

Chemistry
1 answer:
Mars2501 [29]4 years ago
8 0
AG is the element solver
You might be interested in
Can someone pls help
zhannawk [14.2K]

Answer:

1.2km=1,200,000! hope this helps

4 0
3 years ago
When you standardized the Na2S2O3, what molarity of Na2S2O3 did you obtain?
nikdorinn [45]

Answer:

0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)

Explanation:

Based on the reaction:

6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI

<em>6 moles of  Na₂S₂O₃ react per mole of KIO₃</em>

Assuming the molarity of the KIO₃ solution is 0,1M:

Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = <em>5.0x10⁻⁴ moles</em>

As 6 moles of thiosulfate reacted per mole of iodate:

5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ /  1 mole KIO₃) =

<em>3.0x10⁻³ moles of Na₂S₂O₃. </em>In 6.5mL (6.5x10⁻³L):

3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃

6 0
3 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
4 years ago
Which of the following are in the human brain?
mezya [45]
The correct answer is D :)
5 0
3 years ago
Read 2 more answers
A student was conducting an experiment in the laboratory that had a temperature of 272 °C. The student needed to log the tempera
lozanna [386]

Answer:

d 545K

Explanation:

0°C = 273K

so, 272°C = (273+272)K

8 0
2 years ago
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