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xxMikexx
3 months ago
6

Which of the following numerical expressions gives the number of particles in 2.0 g of Ne?

Chemistry
1 answer:
KiRa [2.9K]3 months ago
7 0
To find the number of particles, use the formula: 2.0 g multiplied by (6.0 x 10^23 particles per mol) divided by 20.18 g/mol. Thus, Option C is accurate.
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If three potatoes have a mass of 667 g, what will be the mass of 100 potatoes? A. 200 kg B. 22.2 kg C. 2223 g D. 20.0 kg
alisha [2963]

Response: B- 22.2 kg

Explanation: Given that three potatoes weigh 667 g, it's implied that one potato weighs 667/3= 222.33 g (approximately), leading to the conclusion for 100 potatoes being 100*222.33= 22233 g, which converts to 22.2 kg since 1 g=1000 kg

8 0
2 months ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
Tems11 [2777]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

Where P represents the pressure of each component at equilibrium.

Starting with an initial pressure of H2S at 3.00 atm, the equilibrium concentrations are:

H2S = 3.00 atm - 2X

H2 = 2X

S2 = X

Substituting these values into the equation gives:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Calculating X yields:

X = 0.008945 atm

In equilibrium, the pressure of S2 is X, so the pressure stands at 0.008945 atm

7 0
1 month ago
A 10.00 g sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 g BaSO4 (M- 233.4). W
eduard [2782]

Answer:

The salt identified is barium chloride.

Explanation:

BaX_2++Na_2SO_4\rightarrow BaSO_4+2NaX

The moles of barium sulfate produced are \frac{11.21 g}{233.38 g/mol}=0.0480 mol

per the reaction, 1 mole of barium sulfate arises from 1 mole of BaX_2.

Therefore, 0.0480 moles result from:

\frac{1}{1}\times 0.0480 mol=0.0480 mol of BaX_2.

The quantity of BaX_2 used amounts to 10.00 g

Moles of BaX_2 = \frac{10.00 g}{\text{Molar mass}}[/tex]

0.0480 mol=\frac{10.00}{\text{Molar mass}}

The molar mass of BaX_2 is 208.33 g/mol

The closest answer to our calculation is BaCl_2=208.2 g/mol.

The correct identification is barium chloride, which has a molar mass of 208.2 g/mol.

8 0
2 months ago
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