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3 years ago

What is the unit of measurement used to express density

Chemistry

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Density Grams or kilograms
Volume is liters milliliters or cm cubed

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What is the molarity of 5.63 moles of lithium in 3.25 liters of solution

luda_lava [24]

Answer:

1.73 Molar

Explanation:

The formula is Molarity=moles of solute/liters of solution, which can be written in whatever way you prefer, and examples include: M=N/V or M=mol/L.

M=N/V

M= $\frac{5.63}{3.25}$

Divide 5.63 by 3.25. When you calculate this, you get 1.73, therefore your answer is 1.73 molar.

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3 years ago

How many kilograms are 1.21 x 10^-6 moles of zinc ?

Svetradugi [14.3K]

Answer: The answer is 2.3 x 10 4's.

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3 years ago

What major gases are absorb by the ocean

zysi [14]

Answer:

C) carbon dioxide and hydrigen

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3 years ago

Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if

PIT_PIT [208]

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

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3 years ago

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

3 years ago