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Luda [366]
2 years ago
6

you just got home from a run on a hot Atlanta afternoon. you grab a 1.00-liter bottle of water and drink three-quarters of it in

one swig. How many moles of water did you consume?
Chemistry
1 answer:
nikitadnepr [17]2 years ago
3 0

Answer:

41.67 mol

Explanation:

1 Litre of water = 1000g

Mole = mass / molar mass

Mass of 1 L of water = 1000 g

Molar mass of water (H2O) :

(H = 1, O = 16)

H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol

Amount of water consumed = (3/4) of 1 litre

= (3/4) * 1000g

= 750g

Therefore mass of water consumed = 750g

Mole = 750g / 18g/mol

Mole of water consumed = 41.6666

= 41.67 mol

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If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 196 N B. 32.67 N End of exam C. 1.96 kg
garri49 [273]
Force = mass x gravity

Force = 20 kg x 9.8 m/s²

Force = 196 Newtons

Answer A

hope this helps!

7 0
3 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
Atoms full valence electron shells are chemically stable. Most noble gases have eight valence electrons. So, one bottle for chem
kap26 [50]
Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.
6 0
2 years ago
Read 2 more answers
In which layer of Earth’s interior does convection occur?
lianna [129]

Answer:

The mantel. I'm pretty sure.

Explanation:

8 0
3 years ago
What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0
3 years ago
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