CHEMISTRY MIDDLE SCHOOL

A liquid that occupies a volume of 0.820 L has a mass of 2.56 kg. What is the density of this liquid in kg/L?

Answers

Answer 1
Answer:

The density of this liquid is 0.320 kg/L.

What is density?

Density is the mass occupied by the substance in unit volume. This density is essential in determining whether the substance floats or sinks.

The equation for density is:

Density= mass÷volume

Volume of the Liquid =  0.820 L

Mass of the liquid  = 2.56 kg.

Density= mass÷volume

Density= 2.56 kg÷0.820 L

Density= 3.12195122

Hence, 3.12195122 is the density of this liquid in kg/L.

Learn more about density here:

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Answer 2
Answer:

Answer: 3.12195

Explanation:

D = M/V


Related Questions

COLLEGE

Use the IUPAC nomenclature rules to give the name for this compound - N2O3.

Answers

Ote: some of these materials have more than one name.

1 carbon dioxide
2 carbon monoxide
3 sulfur dioxide
4 sufur trioxide
5 nitrous oxide
6 nitric oxide
7 dinitrigen trioxide
8 nitrogen dioxide
9 dinitrogen tetroxide
10 dinitrogen pentoxode
11 phosporous trichloride
12 phosphorous pentachloride
13 ammonia
14 sulfur hexachloride
15 phosphorous pentoxide (phosporous V oxide)
16 carbon tetrachloride
17 silicon dioxide
18 carbon disulfide
19 oxygen difluoride
20 PBr2 (I've never heard of this; do you mean PBr3?)

PBr2 would be phosphorous dibromide PBr3 would be phosphorous tribromide
I think the answer is N2O³ BUT I'm not sure
MIDDLE SCHOOL

How many moles of H2O will be produced by 7.37moles of N2H4?
N2H4 + 2H202 -----> N2 + 4H20

Answers

Answer:

n H2O = 29.48 mol

Explanation:

  • N2H4 + 2H2O2 → N2 + 4H2O

⇒ n H2O = ( 7.37 mol N2H4 )( 4 mol H2O / mol N2H4 )

⇒ n H2O = 29.48 mol

HIGH SCHOOL

Suppose you wanted to find out how many milliliters of 1.0 m agno3 are needed to provide 169.9 g of pure agno3? what is step 1 in solving the problem? calculate moles agno3 needed what is the molar mass of agno3? 169.87 g/mol how many milliliters of solution are needed?

Answers

M(AgNO₃) = 169.9 g; mass.
n(AgNO₃) = m(AgNO₃) ÷ M(AgNO₃).
n(AgNO₃) = 169.9 g ÷ 169.87 g/mol.
n(AgNO₃) = 1 mol; amount of silver nitrate.
c(AgNO₃) = 1 M = 1 mol/L.
V(AgNO₃) = n(AgNO₃) ÷ c(AgNO₃).
V(AgNO₃) = 1 mol ÷ 1 mol/L.
V(AgNO₃) = 1 L.
V(AgNO₃) = 1 L · 1000 mL/L.
V(AgNO₃) = 1000 mL; volume of silver nitrate.


HIGH SCHOOL

What is the volume 0.0023 moles of CO2

Answers

1 mole ---------------- 22.4 ( at STP )
0.0023 moles -------- ?

v = 0.0023 * 22.4 / 1

v = 0.05152 L

hope this helps!
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