200 ml of wagter increases to 203.5 ml what is the volume in milliliters

Number of moles in 31.99g O2 <br> And 176.8g NaCl

sergey [27]

#O_2

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{31.99}{32}$

$\\ \sf\longmapsto No\:of\:moles\approx 1$

#NaCl

$\\ \sf\longmapsto No\:of\:moles=\dfrac{176.8}{58.5}$

$\\ \sf\longmapsto No\;of\:moles=3mol$

8 0

2 years ago

Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.(c) SF₆(g), SF₄(g)

Andre45 [30]

The increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

SF₄(g) has lower standard molar entropy. Due to less complexity of this molecules.

While, complexity increases from SF₆(g) to S₂F₁₀(g). Therefore, the standard molar entropy of S₂F₁₀(g) is greater than SF₆(g).

Thus, we concluded that the increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

learn more about standard molar entropy:

brainly.com/question/15908262

#SPJ4

7 0

2 years ago

A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of

blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol$

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

0.01 mol of $H^+$ ion will react with = $\frac{1}{1}\times 0.01mole$ of $OH^-$ ion

Thus (0.012-0.01)= 0.002 moles of $OH^-$ are left in 20 ml or 0.02 L of solution.

$[OH^-]=\frac{0.002}{0.02L}=0.1M$

$pOH=-log[OH^-]$

$pOH=-log[0.1]=1$

$pH+pOH=14$

$pH=14-1=13$

Thus the pH of resulting solution will be 13

7 0

3 years ago

What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?

choli [55]

Answer:

$M=0.638M$

Explanation:

Hello!

In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

$n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol$

Then, the liters of solution:

$V=325mL*\frac{1L}{1000mL} =0.325L$

Finally, the molarity turns out:

$M=\frac{0.207mol}{0.325L}\\\\M=0.638M$

Best regards!

5 0

3 years ago

Predict whether each of the following is soluble in water or not.

wlad13 [49]

Answer:

The soluble in water are a and b. c is not soluble in water

Explanation:

The water solubility of a compound, it depends its polarity (. KHSO4 has several polar molecules, actually it has ions charged positively like K and H, and charged negatively like sulfate (SO4-2). When it happens, molecules can do hydrogen bridges with water. It is because of above, KHSO4 can be used in wine making. Something similar happens with propylen glycol but in this case the polar part is OH- ions. Benzene has not polar parts. It is totally apolar compound.

I hope my answer helps you

4 0

4 years ago