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Paha777 [63]
3 years ago
13

In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury

Chemistry
1 answer:
Nuetrik [128]3 years ago
5 0

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS  = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211  = 0.84789 g

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ELEN [110]

Answer:

1. went on the first manned launch during the Apollo mission

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3. 4500 hour

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4 0
3 years ago
Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl
frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is 2.29\times 10^{4} gal.

Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0
3 years ago
Will mark brainliest. Anyone willing to do this?
Norma-Jean [14]

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ACTIVITY 1

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7 0
3 years ago
Q 9 The table lists the steps to clean water for drinking purpose. 1. Adding chlorine tablets to the water.2. Pouring the water
love history [14]

Answer:

1 and 2

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8 0
2 years ago
On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
3 years ago
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