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9966 [12]
2 years ago
9

Element X is found in two forms: 90.0% is an isotope that has a mass of 20.0, and 10.0% is an isotope that has a mass of 22.0. W

hat is the atomic mass of element X?
A. 20.0
B. 20.2
C. 20.8
D. 21.2
E. 21.8
Can anyone explain this one for me?
Chemistry
1 answer:
VMariaS [17]2 years ago
7 0
M_{X} = \frac{(20.0\cdot 90\%)+(22.0\cdot10\%)}{100\%} = \frac{2020}{100} = 20.2[u]

answer: B
You might be interested in
A precipitate of zinc hydroxide can be formed using the reaction below.
worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH  = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0
2 years ago
CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)
REY [17]
Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

6 0
2 years ago
In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used
DENIUS [597]

Answer:

A. Water and Sugar  can be separated by evaporation and then crystallization

B. Mixture of Hexane and Octane can be separated by distillation

C. Solid Iodine, I₂ and NaCl  can be separated by filtration and then evaporation

D. "Sharpie" permanent marking pen  can be separated by  chromatography

E. Nickel shavings and copper pellets can be separated by magnetic separation

Explanation:

A. A mixture of water and sugar can be separated by employing two separation techniques, evaporation and crystallization. First the sugar solution  is heated to evaporate most of the water. When the solution becomes very saturated, it is allowed to cool and then the sugar molecules are obtained through crystallization induced by seeding or scratching the walls of the container.

B. A mixture of hexane (boiling point = 68 °C) and Octane (boiling point = 125 °C) can be separated by distillation due to their significant difference in boiling points.

The mixture is heated in a flask connected to a Liebig condenser. Hexane with the lower boiling point will distill over first and is collected. Afterwards, octane next distills over and is collected as well.

C. A mixture of solid iodine and NaCl can be seperated by first dissolving in water. Iodine being non- polar does not dissolve and is collected as a residue from filtration using a filter paper, while the NaCl solution is collected as the filtrate. The NaCl is recovered from solution by evaporating to dryness in an evaporating dish.

D. "Sharpie" permanent marking pen contains a mixture of dyes which can be separated by paper chromatography.

A drop of the marker ink is placed on a spot above the solvent level on the paper strip used for the separation. The paper strip is  held vertically inside a jar containing a solvent which serves as the mobile phase. The jar is covered and the different dyes move along the paper which serves as the stationary phase, and is thus separated. The paper strip is removed from the jar when the ascending front of the solvent is approaching the top of the paper. The paper is dried and the various dyes can be identified by comparing the distance each has traveled with those of standards.

E. A mixture of nickel shavings and copper pellets can be separated by magnetic separation.

A magnet is brought near the mixture and the nickel shavings being magnetic is attracted to the magnet leaving copper pellets behind since copper is not magnetic.

4 0
2 years ago
Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1
erastovalidia [21]

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, <u>in this case</u> / Total mass) . 100

<u>We were asked for the acetonitrile</u> → (667.7 g / 824 g) . 100 = 81%

3 0
2 years ago
A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s
Mama L [17]
Use the formula, Q= mcT

Q= heat
m= mass= 1.900Kg= 1.900 x 10^3 grams
c= specific heat= 3.21 
T= 4.542 K

Q= (1.900 x10^3g)(3.21)(4.542K)= 14.6 Joules.
7 0
2 years ago
Read 2 more answers
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