What is the chemical formula for calcium hydroxide and explain why

In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00kg o

Lerok [7]

M = amount of the solute / mass of the <span>solvent

0.523 = x / 2.00

x = 0.523 * 2.00

x = 1,046 moles

molar mass KI = </span><span>166.0028 g/mol
</span><span>
Mass = 1,046 * 166.0028

Mass </span>≈<span> 173.63 g

hope this helps!

</span>

6 0

3 years ago

What would happen if the Earth was tilted more than its current tilt of 23.5°? <br><br> HELP ASAP!!

Oduvanchick [21]

Answer:

Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?

Explanation:

Do NOT trust me.

7 0

3 years ago

If you have a 4.6 L of gas in a piston at a pressure of 1.2 atm and compress the gas unit it's volume is 2.6 L, what will the ne

klio [65]

Answer:

<h3>The answer is 2.12 atm</h3>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

So we have

$P_2 = \frac{4.6 \times 1.2}{2.6} = \frac{5.52}{2.6} \\ = 2.1230769...$

We have the final answer as

Hope this helps you

8 0

3 years ago

Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a

nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

$A->B;\Delta _rG^o=-13 kJ/mol$

$C ->D ;\Delta _rG^o=3.5 kJ/mol$

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

$Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167$

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

$K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249$

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

6 0

3 years ago

Determine the concentrations of BaBr2, Ba2 , and Br– in a solution prepared by dissolving 1.18 × 10–4 g BaBr2 in 1.00 L of water

Anika [276]

Supposing complete ionization:
<span>BaBr2 → Ba{2+} + 2 Br{-} </span>

<span>(2.23 × 10^–4 g BaBr2) / (297.135 g BaBr2/mol) / (2.00 L) = 3.75 × 10^-7 mol/L BaBr2 </span>

<span>(3.75 × 10^-7 mol/L BaBr2) x (1 mol Ba{2+} / 1 mol BaBr2) = 3.75 × 10^-7 mol/L Ba{2+} </span>

<span>(3.75 × 10^-7 mol/L BaBr2) x (2 mol Br(-} / 1 mol BaBr2) = 7.50 × 10^-7 mol/L Br{-}</span>

5 0

3 years ago