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3 years ago

Which of the samples pictured is most likely a nonmetal?

Chemistry

1 answer:

max2010maxim [7]3 years ago

7 0

Answer:

you need to be more specific

try adding a picture of the samples

Explanation:

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The salts NaCl and CaCl

Irina18 [472]

The answer is 8 :)
All nobel gases have 8 outer electrons.

6 0

3 years ago

A 687.80-g sample of a noble is held in a 15 L cylinder tank at 3800 torr and a temperature of 22 degrees C. Identify the gas.

BARSIC [14]

Answer:.......................

8 0

3 years ago

Two different compounds are obtained by combining nitrogen with oxygen. the first compound results from combining 46.7 g of n wi

kondaur [170]

First:

46.7 g of N with 53.3 g of O,

=> mass ratio O to N = 53.3 / 46.7 = 1.1413

Second

17.9 g of N and 82.0 g of O.

mass ratio of O to N = 82.0 / 17.9 = 4.5810

Third

Ratio of the mass ratio of O to N in the second compound to the mass ratio of O to N in the first compound =

= 4.5810 / 1.1413 = 4.013 ≈ 4

Answer: 4

8 0

3 years ago

This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press

Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀ = - kt

thus:

(pPH₃ )t = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0

3 years ago

Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of

Natalija [7]

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = $\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}$

First we calculate the initial concentration:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We put the data in the Kc expression and solve for x:

$\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}$

$\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}$

We make a simplification because x<<< 0.0225:

$\frac{4x^3}{0.0225} =9.30*10^{-8}$

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

5 0

3 years ago

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