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4 years ago

PLEASE HELP!!!

Chemistry

1 answer:

sergij07 [2.7K]4 years ago

4 0

oxygen, hydrogen, sulfur, iodine, iron, carbon, nitrogen, phosphorus, calcium.

are the answers

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Which atom attains a stable valence electron configuration by bonding with another atom?

zloy xaker [14]

Answer:hydrogen .

Explanation:

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3 years ago

What is the frequency of a wave of intrared light which has a wavelength of 4.35 * 10 ^ - 5 meters?

weqwewe [10]

Answer

6.9 x 10¹²Hz

Explanation:

Given parameters:

Wavelength of the wave = 4.35 x 10⁻⁵m

Unknown:

Frequency of the wave = ?

Solution:

The speed of a wave is related to the wavelength and frequency using the expression below;

C = Fλ

C is the speed of light = 3 x 10⁸m/s

F is the frequency

λ is the wavelength

Insert the given parameters and solve;

3 x 10⁸ = F x 4.35 x 10⁻⁵

F = $\frac{3 x 10^{8} }{4.35 x 10^{-5} }$ = 6.9 x 10¹²Hz

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3 years ago

The value of ksp for silver sulfide, ag2s, is 8.00×10−51. calculate the solubility of ag2s in grams per liter.

Ivanshal [37]

When Ag₂S dissolves, it dissociates as follows;
Ag₂S ---> 2Ag⁺ + S²⁻
First we need to calculate molar solubility which gives the number of moles dissolved in 1 L of solution.
If molar solubility of Ag₂S is y, then molar solubility of Ag²⁺ and S²⁻ is 2y and y respectively.
ksp gives the solubility constant
ksp = [Ag⁺]²[S²⁻]
ksp = [2y]²[y]
4y³ = 8.00 x 10⁻⁵¹
y³ = 2 x 10⁻⁵¹
y = 1.26 x 10⁻¹⁷ mol/L
molar mass = 247.8 g/mol
solubility of Ag₂S = 1.26 x 10⁻¹⁷ mol/L x 247.8 g/mol = 3.12 x 10⁻¹⁵ g/L
Solubility of Ag₂S = 3.12 x 10⁻¹⁵ g/L

7 0

4 years ago

Read 2 more answers

The percentage of Chlorine in the sample

seropon [69]

Im sorry really dont know the answer

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3 years ago

. Monthly measurements of atmospheric CO2 concentration at Mauna Loa began in March 1958. The average CO2 concentration for that

Harrizon [31]

Answer:

$Increase=98.8ppm$

$Average\ increase/year=1.594\frac{ppm}{year}$

Explanation:

Hello,

In this case, since the nowadays concentration of CO2 is 414.51 ppm and the concentration in 1958 was 315.71 ppm, the total increase is computed via the difference between them:

$Increase=414.51ppm-315.71ppm\\\\Increase=98.8ppm$

Moreover, the average increase per year is computed considering that from 1958 to 2020, 62 years have passed, therefore, such average is:

$Average\ increase/year=\frac{98.8ppm}{62 years} \\\\Average\ increase/year=1.594\frac{ppm}{year}$

Regards.

4 0

3 years ago