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Delvig [45]
2 years ago
8

Nitrogen dioxide and water react to produce nitric acid, HNO3,

Chemistry
1 answer:
maw [93]2 years ago
7 0

Answer:

Explanation:

a )

3NO₂(g) + H₂O(l) — -→ 2HNO₃(aq) + NO(g)

3 x 46 g       18 g            2 x 63 g       30 g

138 g of NO₂ requires 18 g of H₂O

28 g of NO₂ requires ( 18 / 138) x 28

= 3.65 g of H₂O.

b )

18 g of H₂O produces 30 g of NO gas

15.8 g of H₂O produces ( 30/18 ) x 15.8

= 26.33 g of NO gas .

c )

138 g of NO₂ produces 126 g of HNO₃

8.25 g  of NO₂ produces (126 / 138 ) x 8.25

= 7.53 g of HNO₃

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How is the process of mitosis essential to the survival of an organism?
ELEN [110]

Answer:

Explanation:

Mitosis is crucial to this process. Mitosis is the reason we can grow, heal wounds, and replace damaged cells. Mitosis is also important in organisms which reproduce asexually: this is the only way that these cells can reproduce. This is the one key process that sustains populations of asexual organisms.Jul 22, 2020

3 0
2 years ago
Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no
nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True.  Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1

for example, for a reversible reaction  ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)

D) True. Standard conditions refer to

T= 298 K

pH= 7

P= 1 atm

C= 1 M for all reactants

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5 0
3 years ago
The molecular weight of a gas that has a density of 5.75 g/l at stp is __________ g/mol.
Ira Lisetskai [31]

Answer : The molecular weight of a gas is, 128.9 g/mole

Explanation : Given,

Density of a gas = 5.75 g/L

First we have to calculate the moles of gas.

At STP,

As, 22.4 liter volume of gas present in 1 mole of gas

So, 1 liter volume of gas present in \frac{1}{22.4}=0.0446 mole of gas

Now we have to calculate the molecular weight of a gas.

Formula used :

\text{Moles of gas}=\frac{\text{Mass of a gas}}{\text{Molecular weight of a gas}}

Now put all the given values in this formula, we get the molecular weight of a gas.

0.0446mole=\frac{5.75g}{\text{Molecular weight of a gas}}

\text{Molecular weight of a gas}=128.9g/mole

Therefore, the molecular weight of a gas is, 128.9 g/mole

8 0
3 years ago
Read 2 more answers
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Answer:

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IV. Rate of reaction increases due to increase temperature favouring both directions of the equilibrium - causing products to form faster.

Hope this helps!

6 0
2 years ago
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Stells [14]
False is the answer



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5 0
3 years ago
Read 2 more answers
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