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2 years ago

In volumetric analysis, how does one convert concentration from g/dm3 to mol/dm3?

Chemistry

1 answer:

Cloud [144]2 years ago

7 0

gdm^-3/gmol^-1

g/g=1 dm^-3/1=dm^3 1/mol^-1=mol^1 1 x dm^-3 x mol^1 mol/dm^3

To convert from g/dm^3 to mol/dm^3 you have to divide g/dm^3 by g/mol or the molecular mass.

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Given that:2NO (g)→N2 (g)+O2 (g)N2 (g)+O2 (g)+Cl2 (g)→2NOCl (g) has an enthalpy change of −180.6 kJ has an enthalpy change of 10

likoan [24]

Answer:

Explanation:

It is possible to answer this question knowing Hess's law that says you can sum half-reactions enthalpy cahnge to obtain enthalpy change of the total reaction. Using the reactions:

(1) 2NO(g) → N₂(g)+O₂(g) ΔH = -180,6 kJ

(2) N₂(g) + O₂(g) + Cl₂(g) → 2NOCl(g) ΔH = +103,4 kJ

The reverse reactions of (1) and (2) are:

N₂(g)+O₂(g) → 2NO(g) ΔH = +180,6 kJ

2NOCl(g) → N₂(g) + O₂(g) + Cl₂(g) ΔH = -103,4 kJ

The sum of these reactions is:

2NOCl(g) → 2NO(g) + Cl₂(g) ΔH = +180,6 kJ -103,4 kJ = 77,2 kJ

I hope it helps!

6 0

2 years ago

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the pe

ad-work [718]

We know the law of conservation of mass

It states that mass is neither formed nor destroyed in any chemical reaction.
Mass of reactants=Mass of products.

Here

Mg and I_2 are reactants
MgI_2 is product with some yield.
Mass of reactants=10+60.0=70.0g
Mass of MgI_2=53.88g
Mass of yield=Product-MgI_2=70-53.88=16.12g

Lets find the percentage

$\\ \tt\hookrightarrow \dfrac{Mass\:of\:yield}{Total\:mass}\times 100$

$\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100$

$\\ \tt\hookrightarrow 0.23028(010)=23.028\%$

6 0

2 years ago

Read 2 more answers

What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container?

AleksandrR [38]

Answer:

The answer is C. 7.52 K

Explanation:

hope it helps

6 0

2 years ago

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

melisa1 [442]

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

$q = m Cwater$ ΔT $q = m (4.184\frac{J}{(g.^{0}C)} )$ $(0^{0}C-50^{0}C)$

Calculating heat absorbed by 16 g of ice: Ice at $-12^{0}C$ is converted to ice at $O^{0}C$ and then ice at $O^{0} C$ to water at $0^{0}C$

$q = m Cice$ ΔT + $m (Heat of fusion)$

$q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C))$ + $16 g (333.55\frac{J}{g})$

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-( $m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J$

m = 27.4 g

Therefore, 27.4 g water at $50^{0}C$ must be added to 16 g of ice at $-12^{0}C$ to convert to liquid water at $0^{0}C$

4 0

3 years ago

Picture attached, $3

Luba_88 [7]

Answer:

its -000.5

Explanation:

6 0

3 years ago