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LiRa [457]
3 years ago
6

A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an

d liquid combined have a mass of 120.5 g. Calculate the density of the liquid in g/cm3.
Chemistry
1 answer:
Delvig [45]3 years ago
3 0

<u>Answer:</u> The density of liquid is 1.12g/cm^3

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, m_2 = (M-m_1)=(120.5-65.1)g=55.4g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = 49.3 cm^3    (Conversion factor:  1mL=1cm^3 )

Putting values in above equation, we get:

\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3

Hence, the density of liquid is 1.12g/cm^3

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Helium

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Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

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2) Add a solution of NaBr

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Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr

PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr

Therefore, the best solution for dissolving PbBr₂(s) is NaBr

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Answer: Significant figures in a measurement are all measured digits, and one estimated digit

Significant figures communicate the level of precision in measurements Significant figures are an indicator of the certainty in measurements.

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Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy or precision is known as significant digits.

The significant figures of a measured quantity are defined as all the digits known with certainty and the first uncertain or estimated digit.

Rules for significant figures:

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2. All non-zero numbers are always significant.

3. All zero’s between integers are always significant.

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