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mixer [17]
3 years ago
7

What happens when potassium forms an ion?

Chemistry
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

It loses an electron and has an octet in the next-lowest energy level

Explanation:

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Diano4ka-milaya [45]
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3 years ago
Review Toples
jekas [21]

Answer:

1.553 M^2s^-1

Explanation:

Now let us consider experiments 1,2,3.

In experiment 1 and 2, the concentration of NO is doubled and the initial rate of reaction is quadrupled.

0.575/0.144 = 4

So;

2^2 =4

In experiment 2 and 3

The concentration of reaction of H2 is doubled and the rate of reaction is doubled.

0.575/0.287 = 2

2^1 =2

So we can now write;

Rate =k[NO]^2 [H2]^1

Taking up the first reaction;

0.144 =k [0.475]^2 [0.411]

k = 0.144/ [0.475]^2 [0.411]

k = 1.553 M^2s^-1

5 0
3 years ago
What does it mean when a rock is foliares?
yan [13]
Started out as some other type of rock, but have been substantially changed from their original igneous, sedimentary, or earlier metamorphic form. Metamorphic rocks form when rocks are subjected to high heat, high pressure, hot mineral-rich fluids or, more commonly, some combination of these factors.
6 0
3 years ago
Read 2 more answers
Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2
Elena-2011 [213]

Answer:

Approximately 0.224\;\rm L, assuming that this reaction took place under standard temperature and pressure, and that \rm CO_2 behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that \rm CaCO_3 is the limiting reactant. How many moles of \rm CO_2 would be produced?

Look up the relative atomic mass of \rm Ca, \rm C, and \rm O on a modern periodic table:

  • \rm Ca: 40.078.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass of \rm CaCO_3:

\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the number of moles of formula units in 1\; \rm g of \rm CaCO_3 using its formula mass:

\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}.

In the balanced chemical equation, the ratio between the coefficient of \rm CaCO_3 and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1.

In other words, for each mole of \rm CaCO_3 formula units consumed, one mole of \rm CO_2 would be produced.

If \rm CaCO_3 is indeed the limiting reactant, all that approximately 1.00\times 10^{-2}\; \rm mol of \rm CaCO_3\! formula would be consumed. That would produce approximately 1.00\times 10^{-2}\; \rm mol\! of \rm CO_2.

On the other hand, assume that \rm HCl is the limiting reactant.

Convert the volume of \rm HCl to \rm dm^{3} (so as to match the unit of concentration.)

\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 5.00\times 10^{-2}\; \rm dm^{3} of this \rm 0.05\; \rm mol \cdot dm^{-3}

\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}.

Notice that in the balanced chemical reaction, the ratio between the coefficient of \rm HCl and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}.

In other words, each mole of \rm HCl molecules consumed would produce only 0.5\;\rm mol of \rm CO_2 molecules.

Therefore, if \rm HCl is the limiting reactant, that 2.50 \times 10^{-3}\; \rm mol of \rm HCl\! molecules would produce only one-half as many (that is, 1.25\times 10^{-3}\; \rm mol) of \rm CO_2 molecules.

If \rm CaCO_3 is the limiting reactant, \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. However, if \rm HCl is the limiting reactant, 1.25\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be produced.

In reality, no more than \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. The reason is that all \rm CaCO_3 would have been consumed before \rm HCl was.

After finding the limiting reactant, approximate the volume of the \rm CO_2\! produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately 22.4\; \rm L.

If \rm CO_2 behaves like an ideal gas, the volume of that \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be approximately \rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L.

3 0
3 years ago
What independent variables can be tested in this lab setup? (List at least 5)
tigry1 [53]

Answer:

I know 2

dependent

independent

Explanation:

sorry that's about it from me

6 0
3 years ago
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