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slega [8]
3 years ago
10

How many millimeters are there in 1 kilometer

Chemistry
1 answer:
Marysya12 [62]3 years ago
4 0
1.000.000 is the correct answer
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We want to find how much charge is on the electrons in a nickel coin. follow this method. a nickel coin has a mass of about 5 g.
Lady bird [3.3K]
The mass of a nickel coin is 5 g.
1 mol of Ni weighs 58 g. 1 mol contains 6.022 x 10²³ atoms of Ni.
therefore in 58 g  there are 6.022 x 10²³ atoms of Ni
then in 5 g the number of Ni atoms are - 6.022 x 10²³ /58 x 5 = 5.2 x 10²² Ni atoms
Therefore number of Ni atoms are 5.2 x 10²² atoms in a nickel coin

3 0
3 years ago
Bases feel slippery because they remove fats and oils from the surface of our skin.
cluponka [151]
It is true hope it helps!
6 0
3 years ago
Read 2 more answers
The water-gas shift reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is used industrially to produce hydrogen. The reaction enthalpy is ΔH∘=−4
denis23 [38]

Answer:

To increase the yield of H₂ we would use a low temperature.

For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.

Explanation:

Let´s consider the following reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.

In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.

3 0
3 years ago
For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?
gogolik [260]
By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6
3 0
3 years ago
Read 2 more answers
Can somebody help me, please!
zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 72/100= 0.72      or        72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

  • mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 28/100= 0.28       or        28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

  • mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

<u>Atomic Mass of Rubidium:</u>

Add the two numbers together.

  • Rb-85 (61.2) and Rb-87 (24.36)
  • 61.2+24.36=85.56 amu
4 0
3 years ago
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