All categories

3 years ago

Whatwas the initial volume of the hydrogen in cm3?

Chemistry

2 answers:

svetlana [45]3 years ago

5 0

Answer:

255.51cm3

Explanation:

Data obtained from the question include:

V1 (initial volume) =?

T1 (initial temperature) = 50°C = 50 + 273 = 323K

T2 (final temperature) = - 5°C = - 5 + 237 = 268K

V2 (final volume) = 212cm3

Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:

V1/T1 = V2/T2

V1/323 = 212/268

Cross multiply to express in linear form

V1 x 268 = 323 x 212

Divide both side by 268

V1 = (323 x 212)/268

V1 = 255.51cm3

Therefore, the initial volume of the gas is 255.51cm3

bija089 [108]3 years ago

4 0

Answer:

Explanation:

To get the initial volume of the hydrogen in cm³, we apply the equation generated from Charles' law.

V1/T1=V2/T2

Where;

V1=initial volume of gas

T1=initial temperature of gas

V2=final volume of gas

T2=final temperature of gas

We make V1 subject of formula

V1=(V2×T1)/T2

Given;

V2=212cm³

T1=50°C=(50+273)K=323K

T2=-5.0°C=(-5.0+273)K=268K

V1=(212×323)/268

V1=255.5cm³

Therefore, the initial volume of the hydrogen was 255.5cm³

You might be interested in

The correct mathematical expression for finding the molar solubility (s) of calcium phosphate is

GalinKa [24]

Answer is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.

Balanced chemical reaction: Ca₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).
[Ca²⁺] = 3s(Ca₃(PO₄)₂) = 3s.
[PO₄³⁻] = 2s.
Ksp = [Ca²⁺]³ · [PO₄³⁻]².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.

s = ⁵√(Ksp ÷ 108).

6 0

3 years ago

Your mother has an unlabeled bottle of clear liquid in the refrigerator. You pour a small cup of the liquid and taste it, and it

Nostrana [21]

The pH of the liquid would be 6 or a 5

6 0

3 years ago

Can someone please help me with this?

BabaBlast [244]

Answer:

1. Main sequence stars have different masses. The common characteristic they have is their source of energy. They burn fuel in their core through the process of fusing hydrogen atoms into helium.

2. Supergiants are among the most massive and most luminous stars. Supergiant stars occupy the top region of the Hertzsprung–Russell diagram with absolute visual magnitudes between about −3 and −8. The temperature range of supergiant stars spans from about 3,400 K to over 20,000 K.

3. Supergiants develop when massive main-sequence stars run out of hydrogen in their cores.

4. a supernova occur When the pressure drops low enough in a massive star, gravity suddenly takes over and the star collapses in just seconds. This collapse produces the explosion.

5. when a star has reached the end of its life and explodes in a brilliant burst of light

Explanation:

4 0

3 years ago

Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)

steposvetlana [31]

Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ

6 0

3 years ago

Read 2 more answers

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

$n=\frac{m}{M_m}$