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3 years ago

How will volume affect the mass of a liquid?

Chemistry

1 answer:

erastovalidia [21]3 years ago

5 0

Answer:

it will affect the mass by changing in pressure will affect the volume (and therefore density) of gases, but will not noticeably affect the volume (or density) of solids and liquids. For gases: An increase in pressure leads to a decrease in volume and an increase in density. A decrease in pressure leads to an increase in volume and an decrease in density.

Explanation:

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This is due at 4:00 P.M. and I do not want to make a 76%, so please help MEEEEEEEEEEEEE!!!

Ann [662]

Answer:

Hi, There!...

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Explanation:

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3 years ago

Read 2 more answers

0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

4 0

3 years ago

How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

ozzi

The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below

since M/v% is mass of solute in grams per 100 ml

convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams

volume is therefore = 100 ( mass/ M/v%)

= 100 x( 0.0015/ 0.4) = 0.375 ML

6 0

4 years ago

Use the balanced chemical equation below. How many grams of the product are formed when 2.34 g of sulfur is completely reacted w

levacccp [35]

Answer:

7.89 g

Explanation:

Step 1: Write the balanced equation

S₈ + 16 F₂(g) → 8 SF₄

Step 2: Calculate the moles corresponding to 2.34 g of S₈

The molar mass of S₈ is 256.52 g/mol.

$2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol$