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Simora [160]
2 years ago
5

What feature of the sun changes to create the solar cycle?

Chemistry
1 answer:
Sauron [17]2 years ago
7 0
The magnetic field !
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3. Give three examples of a pure substances
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There will be weight, there will be volume, there will be height

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Which of the following is a nonrenewable resource?
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C. Natural gas

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Arrange the following elements in order of increasing ionization energy
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Ionization Trend: First ionization energy will increase left to right across a period and increase bottom to top of a family (column).

A) Sr, Be, Mg are all in column 2 of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Be, Mg, Sr

B) Bi, Cs, Ba are all in the same row of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Bi, Ba, Cs

C) Same rule as above. Order is: Na, Al, S

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3 years ago
A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec
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<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol

  • <u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol

For the given chemical reaction:

Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, 3.14\times 10^{-5} moles of NaI will react with = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 3.14\times 10^{-5} moles of NaI will produce = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

4 0
3 years ago
in 1912, chemist fritz haber developed a process that combined nitrogen from the air with hydrogen at high temperatures and pres
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Describe how many atoms are involved before and after. what do you notice about the number of atoms?

From the balanced chemical reaction, we see that we need 1 mol of N2 gas and 3 mol of H2 in order to form 2 mol of NH3.
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3 years ago
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