All categories

1 year ago

Please hurry and tell the answer to:

Chemistry

1 answer:

Flura [38]1 year ago

5 0

Answer:

6.50 g of Hydrogen

Explanation:

We know that in every 20.0g of sucrose, there are 1.30g of hydrogen.

We now have 100.0g of sucrose. 100.0g is 5x larger than the 20.0g sample, which is a 5 : 1 ratio. Applying this ratio to the amount of hydrogen, we would have 5*1.3g of hydrogen in the 100.0g of sucrose.

5*1.3 = 6.5, so our answer is that there are 6.50g of hydrogen in 100.0g of sucrose.

Hope this helps!

You might be interested in

21. Draw Lewis structure

Veseljchak [2.6K]

Answer:

C contains one N and three I atoms

7 0

2 years ago

Which line segment on the graph represents boiling

hoa [83]

1. There is no graph given

2. ENERGY LEVELS!!!

Boiling is the process of liquid(middle energy, flat line) turning into a gas(lots of energy, flat line), hope this helps

8 0

3 years ago

SOMEONE PLS HELP

Gekata [30.6K]

Answer:

Explanation:

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

$P_1V_1 = P_2V_2$

Since we're finding the new volume

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

100.7 kPa = 100,700 Pa

95.1 kPa = 95,100 Pa

We have

$V_2 = \frac{100700 \times 439}{95100} = \frac{44207300}{95100} \\ = 464.8506...$

We have the final answer as

Hope this helps you

6 0

2 years ago

At a consent tempture the pressure on a 2.5 l ballon is increased from 2.4 atm to 5.8 atm. what is the new volume?

marissa [1.9K]

Answer:

1.034 L

Explanation:

P1 V1 = P2 V2

P1 V1 / P2 = V2

2.4 (2.5) / 5.8 = V2 = 1.034 L

8 0

1 year ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

5 0

1 year ago