What is the oxidation number of Mn in KMnO₄? a. -7 b. -3 c. 0 d. +3 e. +7

Convert the following into balanced equations. ( Use the lowest possible coefficents)?(a) When gallium metal is heated in oxygen

ehidna [41]

Answer:

Explanation:

(a) Balanced reaction of gallium with oxygen is as follows:

$4Ga(s)+3O_2(g) \rightarrow 2Ga_2O_3 (s)$

(b) $C_6H_{14}(l)+O_2(g)\rightarrow CO_2(g)+H_2O(l)$

Multiply the carbon dioxide by 6 to balance carbon as follows:

$C_6H_{14}(l)+O_2(g)\rightarrow 6CO_2(g)+H_2O(l)$

After that multiply H2O by 7 to balance hydrogen as follows:

$C_6H_{14}(l)+O_2(g)\rightarrow CO_2(g)+7H_2O(l)$

Finally balance oxygen by multiplying O2 by 19/2. Therefore, balanced reaction is as follows:

$C_6H_{14}(l)+\frac{19}{2} O_2(g)\rightarrow 6CO_2(g)+7H_2O(l)$

(c) $Na_3PO_4(s)+CaCl_2 \rightarrow Ca_3(PO_4)_2+NaCl$

first balance calcium, by multiply $CaCl_2$ by 3 as follows:

$Na_3PO_4(s)+3CaCl_2 \rightarrow Ca_3(PO_4)_2+NaCl$

After that balance phosphorous by multiplying $Na_3PO_4$ by 2 as follows:

$2Na_3PO_4(s)+3CaCl_2 \rightarrow Ca_3(PO_4)_2+NaCl$

Finally balance Na by multiplying NaCl by 6. Therefore, balance reaction is as follows:

$2Na_3PO_4(s)+3CaCl_2 \rightarrow Ca_3(PO_4)_2+6NaCl$

3 0

3 years ago

Look at the image shown.

Charra [1.4K]

Where’s the image sorry this isn’t much help but I don’t know what to answer if I can’t see the image

6 0

2 years ago

Polychlorinated biphenyls (PCBs), which were formerly used in the manufacture of electrical transformers, are environmental and

lubasha [3.4K]

C12H4Cl6 + 23 O2 + 2 H2O → 24 CO2 + 12 HCl 

(a) 
(13.5 mol O2) x (2 mol H2O / 23 mol O2) = 1.17 mol H2O 

(b) 
(17.6 mol H2O) x (12 mol HCl / 2 mol H2O) x (36.4611 g HCl/mol) = 3850 g HCl 

(c) 
(90.4 g HCl) / (36.4611 g HCl/mol) x (24 mol CO2 / 12 mol HCl) = 4.96 mol CO2 

(d) 
(106.01 g CO2) / (44.00964 g CO2/mol) x (2 mol C12H4Cl6 / 24 mol CO2) x (360.8782 g C12H4Cl6/mol) = 
72.4 g C12H4Cl6 

(e) 
(4.2 kg C12H4Cl6) / (360.8782 g C12H4Cl6/mol) x (12 mol HCl / 2 mol C12H4Cl6) x (36.4611 g HCl/mol) = 
2.5 kg HCl

4 0

3 years ago

Read 2 more answers

Consider the following reaction: 2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s)

Mandarinka [93]

Taking into account the reaction stoichiometry, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Bi + 3 Cl₂ → 2 BiCl₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Bi: 2 moles
Cl₂: 3 moles
BiCl₃: 2 moles

The molar mass of the compounds is:

Bi: 209 g/mole
Cl₂: 70.90 g/mole
BiCl₃: 315.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Bi: 2 moles ×209 g/mole=418 g

Cl₂: 3 moles ×70.90 g/mole= 212.7 g

BiCl₃: 2 moles ×315.45 g/mole= 630.9 g

<h3>Mass of Cl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 630.9 grams of BiCl₃ are formed by 212.7 grams of Cl₂, 3.32 grams of BiCl₃ are formed by how much mass of Cl₂?

$mass of Cl_{2} =\frac{3.32 grams of BiCl_{3}x212.7 grams of Cl_{2} }{630.9 grams of BiCl_{3}}$