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2 years ago

7

A gas with a volume of 4.0 L at 90.0 kPa expands until the pressure drops to 20.0 kPa. What is its new volume if the temperature

doesn't change?

1 answer:

Flauer [41]2 years ago

4 0

<span>You are trying to find V2, the formula is V2</span>= <span><span>P1 / </span><span>P2 </span></span>⋅ V<span>1
</span><span>input your numbers, V2</span>= <span><span>90.0kPa / </span><span>20.0kPa </span></span>⋅ 4.0 L to get 18 L

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Activity C: Hydrolysis of Carbohydrates As discussed earlier, disaccharides are composed of two monosaccharides linked together.

adelina 88 [10]

Answer:

Sucrose: glucose and fructose

Explanation:

<em>What monosaccharides will result from the hydrolysis of sucrose?</em>

<em>Sucrose</em> is a <em>disaccharide</em> composed of 2 different <em>monosaccharides</em>: glucose and fructose joining by a 1 ⇒ 2 bond. These monosaccharides will be released upon the hydrolysis of sucrose.

<em>What monosaccharide will result from the hydrolysis of starch?</em>

<em>Starch</em> is a <em>polysaccharide</em> composed of numerous glucose monomers joined by glycosidic bonds (1 ⇒ 4 and 1 ⇒ 6). These monosaccharides will be released upon the hydrolysis of starch.

4 0

3 years ago

If 26.2 grams of a pure compound contain 8.77 × 1022 molecules, what is the molecular weight of this compound? Answer in units o

Nataly [62]

Answer:

Mw = 179.845 g/mol

Explanation:

Mw [=] g/mol

∴ w = 26.2 g

∴ 1 mol = 6.02 E23 molecules.......Avogadro's number

⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol

⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol

3 0

2 years ago

You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.

Let's plug in the values in the equation:

$1.25M(363mL)=0.50M(V_2)$

$V_2=\frac{1.25M(363mL)}{0.50M}$

$V_2=907.5mL$

Volume of water added = 907.5mL - 363mL = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0

2 years ago

Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0

3 years ago

A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the pressure is 740.0 mm of mercury. What will it

Zarrin [17]

Answer:

New volume V2 = 92.7 Liter (Approx)

Explanation:

Given:

V1 = 106 l

T1 = 45 + 273.15 = 318.15 K

P1 = 740 mm

T2 = 20 + 273.15 = 293.15 K

P2 = 780 mm

Find:

New volume V2

Computation:

P1V1 / T1 = P2V2 / T2

(740)(106) / (318.15) = (780)(V2) / (293.15)

New volume V2 = 92.7 Liter (Approx)

7 0

2 years ago