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Advocard [28]
3 years ago
8

Is the following equation balanced? SO3 + 2H2O H2SO4 no yes

Chemistry
1 answer:
postnew [5]3 years ago
7 0

Answer: No

Explanation: It's not balanced because four oxygen atoms in H2SO4,  whereas there are 5 oxygen atoms in the reactants side. Also, there's more hydrogen atoms on the reactants side.

I hope this helps!

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To test the purity of sodium bicarbonate, you dissolve a 3.50g sample in water and add sulfuric acid. if 1.04g of carbon dioxide
Andrews [41]

Answer is: the percent purity of the sodium bicarbonate is 56.83 %.

1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.

2. m(NaHCO₃) = 3.50 g

n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).

n(NaHCO₃) = 3.50 g ÷ 84 g/mol.

n(NaHCO₃) = 0.042 mol.

3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.

n(CO₂) = 0.042 mol.

m(CO₂) = 0.042 mol · 44 g/mol.

m(CO₂) = 1.83 g.

4. the percent purity = 1.04 g/1.83 g  ·100%.

the percent purity = 56.8 %.

8 0
3 years ago
Can you help me with three please? We’re balancing electrons
Aneli [31]

Answer:

Explanation:

I did this class yesterday give me like 10min imma find my anwsers

4 0
2 years ago
How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?
mr Goodwill [35]
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
5 0
3 years ago
What is the solute and solvent in roundup
olga nikolaevna [1]

Answer:

solute is that we disolve in solvent

solvent is in which we dissolve solute

8 0
3 years ago
The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at
nalin [4]

Answer:

The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>

Explanation:

The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.

<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.

Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.

Hence , the answer is true .

6 0
2 years ago
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