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alukav5142 [94]
3 years ago
11

If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch

emist have actually prepared? Answer in units of M.
Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

0.120M is the concentration of the solution

Explanation:

<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>

<em />

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

<em>Molar concentration:</em>

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>
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Costs associated with product failures that occur after they have been sold to customers are known as external failure costs. These expenses cover the legal expenditures associated with customer lawsuits.

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One element of the cost of quality is external failure costs, which are incurred when a subpar product is delivered to the client and malfunctions while being used. The warranty work and returns make up the majority of this expense. However, customer lawsuits could also be a possibility.

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1 year ago
El plomo y el yodo forman dos compuestos. En uno, el porcentaje de en masa de plomo es del 44,94% y en el otro es de 62,02%. Cal
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0.8162 gramos de plomo por gramo de yodo

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Así, la masa de plomo por gramos de yodo para el primer compuesto es:

44.94g plomo / 55.06g Yodo =

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<em></em>

Segundo compuesto:

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3 years ago
When the volume of a gas is
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Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

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Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

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