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3 years ago

If the pOH of a solution is 9.3, what is the [OH-]

Chemistry

1 answer:

andriy [413]3 years ago

3 0

Answer:

5.012*10^-10

Explanation:

[OH-] = 10^-pOH

[OH-] = 10^-9.3

[OH-] = 5.012*10^-10

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zmey [24]

Calcium is a non metal

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3 years ago

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The smallest unit of an element that still retains the distinctive behavior of that element is an.

Tju [1.3M]

The smallest unit of an element that still retains the distinctive behavior of that element is an: atom

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A single atom of an element keep its distinctive behavior, that's because the atom is the smallest unit of the composition of a chemical element

The atom is the smallest part of the composition of matter, it is indivisible and is composed of a nucleus that has protons and neutrons, and around the nucleus there are the electrons.

Learn more about the atom at: brainly.com/question/17545314

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2 years ago

A bumblebee helps plants reproduce by carrying pollen from a flower on one plant to a flower on another plant. Which reproductiv

Oliga [24]

Answer:

cross pollination

Explanation:

this is the transfer of pollen grains from the anther of a flower to the stigma of another flower of the same kind.

8 0

3 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the

icang [17]

Answer:

Explanation:

Utilizing Rydber's equation:

ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have

n=1 to n= infinity

ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)

n= 3 to n= infinity

ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J

b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .

1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹

1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m

λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm

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3 years ago

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