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jarptica [38.1K]
3 years ago
6

How many grams of potassium iodide will produce 500 grams of silver iodide, when there is an excess of silver nitrate?Why?

Chemistry
1 answer:
nevsk [136]3 years ago
7 0

Answer:

None of the options are correct. The answer to the question is 353.2g

Explanation:

The equation for the reaction between is given below:

KI + AgNO3 —> KNO3 + AgI

Molar Mass of KI = 39 + 127 = 166g/mol

Molar Mass of AgI = 108 + 127 = 235g/mol

From the equation,

166g of KI produced 235g of AgI.

Therefore, Xg of KI will produce 500g of AgI i.e

Xg of KI = (166 x 500)/235 = 353.2g

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Which of the following correctly lists the name of the element, the symbol for the ion, and the name of the ion?
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A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
3 years ago
216 J of energy is required to raise the temperature of a piece of aluminum from 15.0º C to to 35º C.
ddd [48]

Answer: 12g

Explanation:

The amount of energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 216 joules

Mass of aluminium = ? (let unknown value be Z)

C = 0.90 JºC-1g-1

Φ = (Final temperature - Initial temperature)

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Then, Q = MCΦ

216 J = Z x 0.90 JºC-1g-1 x 20°C

216 J = Z x 18 J°g-1

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Which of the following characteristics helps to distinguish a biome?
ladessa [460]

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Explanation:

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