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Ksenya-84 [330]
3 years ago
15

Which one of the following equations is properly balanced? a. CH 3CHO + 3O 2 → 2CO 2 + 2H 2O b. NH 4NO 3 → 2H 2O + N 2 c. Na 2CO

3 + 2H 2SO 4 → Na 2SO 4 + 2H 2O + CO 2 d. 2Na 2SO 4 + 3Bi(NO 3) 3 → Bi 2(SO 4) 3 + 9NaNO 3 e. Sn + 4HNO 3 → SnO 2 + 4NO 2 + 2H 2O
Chemistry
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

The chemical equation that is properly balanced is;

e. Sn + 4HNO₃ → SnO₂ + 4NO₂ + 2H₂O

Explanation:

To solve the question, we note the reactions thus and we look for a reason why the reaction is not balanced as follows.

a. CH₃CHO + 3O₂ → 2CO₂ + 2H₂O

Here we have 7 oxygen atoms in the reactant and 6 in the products

Not balanced

b. NH₄NO₃ → 2H₂O + N₂

Here we have three oxygen in the reactant and four in the products = Not balanced

c. Na₂CO₃ + 2H₂SO₄ → Na₂SO₄ + 2H₂O + CO₂

Here we have 11 oxygen atoms in the reactant and 8 in the products = Not balanced

d. 2Na₂SO₄ + 3Bi(NO₃)₃ → Bi₂(SO 4)₃ + 9NaNO₃

Here  we have 2 Na in the reactants and 9 Na in the products = Not balanced

e. Sn + 4HNO₃ → SnO₂ + 4NO₂ + 2H₂O

Here we have equal number of products and reactants = Balanced chemcal reaction equation.

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Can a substance be a lewis acid without being a bronsted-lowry acid?argue
storchak [24]

Answer:

Yes

Explanation:

Yes, A substance can be a lewis acid without being a Bronsted-Lowery acid because there are some substances which cannot donate protons(Bronsted-Lowery acid) but can accept a pair of electron.

<u><em>For Example:</em></u>

Let us take the example of BF₃

BF₃ contains no proton so it is not a Bronsted Lowery Acid

However, BF₃ has an incomplete octet with 6 electrons. It needs an electron pair to complete its octet. It accepts a pair of electron to become a Lewis Acid

6 0
3 years ago
Read 2 more answers
What would the molecular formula be if lithium and sulfur reacted to form a neutral compound?
skad [1K]

Answer:

D, Li2S

Explanation:

This is because Lithium, which is in group IA of the periodic table, has a charge of +1. Sulfur will have a charge of -2 because it is in group 6A in the periodic table, which means to balance these out, there needs to be 2 lithium ions which would result in a charge of +2. With Lithium now having a charge of +2 due to having two atoms in the compound, and sulfur already having a charge of -2 as one atom, these two cancel out meaning the compound is neutral.

4 0
3 years ago
What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to
fredd [130]

Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂  →  2KBr + Cr

2mole     1 mole                1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

   ⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

   ⇒ 1.458 x 1/ 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams

6 0
3 years ago
Current Question:
IRISSAK [1]

Answer:

0.2g

Explanation:

All radiodecay follows the 1st order decay equation

A = A₀e^-kt

A => Activity at time (t)

A₀ => Initial Activity at time = 0

k => decay constant for isotope

T => time in units that match the decay constant

Half-Life Equation => kt(½) = 0.693 => k = 0.693/34 min = 0.0204min¹

A = A₀e^-kt  = (26g)e^-(0.0204/min)(238min) = (26g)(0.0078) = 0.203g ~ 0.2g (1 sig fig).

7 0
3 years ago
You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca
defon
The equation to be used are:

PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles

The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa

n = 0.576 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.019862 mol
(101,183.9 Pa)V = (0.019862 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L
5 0
3 years ago
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