PLEASE ANSWER ASSSSPAPPPP

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

Consider the following system at equilibrium:

alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0

3 years ago

a sample of a compound contains 60gC and 5.05gH.it molar mass is 78.12g/mol what is the compound molecular formula

elena55 [62]

(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6

5 0

3 years ago

Which of the following elements has 3 electrons in the outermost d orbital?

Nina [5.8K]

Answer:

A

Explanation:

Element A has 3 electrons in outermost shell so its valency is 3. It will loose 3 electrons to attain stability, as loss of 3 electrons is easier than gain of 5 electrons.

Valency of B is −2 as it will gain 2 electrons to attain stability and combine with other atom.

Valency of A⟶+3

Valency of B⟶−2

(Refer to Image)

Cross multiply valency of A and B

∴A2B3 compound will be formed.

3 0

4 years ago

Look up and record the boiling point of acetic acid, and explain why only some of it evaporates from the reaction mixture.

REY [17]

Answer:

Heating the mixture to a temperature above the boiling point of acetic acid, but below 100°C (the boiling point of water). The vapours from the acetic acid rise, and go into a tube. They are then condensed within the tube, and run off into a separate storage area. Because water can exist as a gas at pretty much any temperature above 0°C, it will result in an impure mixture, but repeatedly doing this will get the acetic acid to the desired purity.

8 0

3 years ago