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mojhsa [17]
3 years ago
9

How many atoms of hydrogen are there in 3.0kg of ethane

Chemistry
1 answer:
siniylev [52]3 years ago
8 0
Vzed Justin c no sas.
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A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of
Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0
3 years ago
You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
How did Thomson show that cathode<br> rays are different from light?
Flura [38]

Answer:

They had a negative charge

Explanation:

5 0
2 years ago
Whats the formula for selenium pentaoxide, iodine trichloride, zinc (1) nitride, chromium (III) bicarbonate, and there molar mas
Phoenix [80]

Answer:

1) Se2O5

2) I2O6

3)Zn3n2

4) Cr(HCO3)3

Explanation:

selenium pentaoxide (= also called diselenium pentoxide)

= Se2O5

⇒ Se = 78.97 g/mol

⇒ O = 16 g/mol

⇒ 2*78.97 + 5*16 = 237.94 g/mol

iodine trichloride

= I2O6

⇒ I = 126.9 g/mol

⇒ Cl = 35.45 g/mol

⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol

zinc (1) nitride does not exist  (it's Zinc(ii)nitride

The oxidation number for zinc is always 2

Zn3n2

⇒ Zn = 65.38 g/mol

⇒ N = 14 g/mol

⇒3*65.38 + 2* 14 = 224.14 g/mol

chromium (III) bicarbonate

Cr(HCO3)3

⇒ Cr = 52 g/mol

⇒ H = 1.01 g/mol

⇒ C = 12 g/mol

⇒ O = 16 g/mol

52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol

3 0
3 years ago
How many total atoms are found in one molecule of alpo4?
Kruka [31]
There is 1 aluminium atom and 4 polonium atoms
5 0
3 years ago
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