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zaharov [31]
3 years ago
10

PLEASE HELP :-)

Chemistry
1 answer:
alex41 [277]3 years ago
6 0
6.4 x 10^-7 = [CO]^2[O2]/ [CO2}^2 = ( 2.0 x 10^-3)^2 ( 1.0 x 10^-3)/ [CO2]^2 = 
<span>=4.0 x 10^-9 / [CO2]^2 </span>

<span>[CO]= sq.rt ( 4.0 x 10^-9)/ 6.4 x 10^-7=7.9 x 10^-2 M </span>


<span>2.6 x 10^-3 = [I]^2 / [I2] = [I]^2 / 0.95 </span>
<span>[I]= sq.rt ( 2.6 x 10^-3 x 0.95)=5.0 x 10^-2 M </span>

<span>Ksp = [Ba2+][CO32-] = ( 1.1 x 10^-4)^2=1.2 x 10^-8</span>
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The precision of a method is being established, and the
Lemur [1.5K]

Answer:

No,  22.09%  is not a valid measurement

Explanation:

Precision has to do with how close a given set of measured values are to each other. It is quite different from accuracy. Accuracy refers to how close a given set of values is to the true value. A given set of values may be precise but not accurate and vice versa.

If we look at the values obtained;  22.09%,  22.15%, 22.18%, 22.23%, 22.25%, the value 22.09% is too far off the other values. This implies that it does not represent a valid measurement since it is not close to all the other values obtained.

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3 years ago
If 185 mg of acetaminophen were obtained from a tablet containing 350 mg of acetamino- phen, what would be the weight percentage
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Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.

Amount of acetaminophen initially taken = 350 mg

Amount of acetaminophen obtained after recovery =185 mg

Weight percentage recovery =\frac{mass recovered}{mass originally taken}*100

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8 0
3 years ago
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Universe.

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5 0
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Read 2 more answers
Determine whether you can swim in 1.00 x 10^27 molecules of water.​
zloy xaker [14]

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

7 0
2 years ago
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