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jek_recluse [69]
3 years ago
14

In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

1.83 g / cm3?
Chemistry
1 answer:
ser-zykov [4K]3 years ago
8 0

Answer:

\large \boxed{\text{183 g}}  

Explanation:

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}

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You will need 0.0264 pounds of baby lotion. How many millimeters of baby lotion will you need if the density of the hand lotion
MatroZZZ [7]

Answer:

v = 10.42 mL

mass = 11.986 g

Explanation:

Given data;

Mass of lotion needed in pound = 0.0264 lb

Density of lotion = 1.15 g/mL

Mass of lotion needed in gram = ?

Volume of lotion needed in mL = ?

Solution:

Mass of lotion needed in gram :

1 pound = 454 g

0.0264 lb × 454 g/ 1lb =  11.986 g

Volume of lotion needed in mL :

d = m/v

1.15 g/mL = 11.986 g / v

v = 11.986 g /1.15 g/mL

v = 10.42 mL

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4 years ago
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Explanation:

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Which components of a galaxy move in circular patterns or revolve around a star?
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3 years ago
What would diluting the vinegar do to the rate of a reaction between vinegar and sodium hydroxide? decrease the rate of reaction
GarryVolchara [31]
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The reaction of vinegar with sodium hydroxide is:

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That is an acid-base reaction.

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Then, given thatn when you dilute the vinegar you decrease the concentration of CH3COOH, the rate of reaction will decrease.

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3 years ago
The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

\frac{7.59 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.367 g

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

\frac{2.11 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.102 g

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

5 0
2 years ago
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