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dolphi86 [110]
3 years ago
11

The percent composition of calcium is ?

Chemistry
1 answer:
DochEvi [55]3 years ago
5 0

Answer: 40.1%

Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.

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How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

4 0
3 years ago
Can someone answer these two questions for me ill give brainliest
gladu [14]
Answer:
1. No
2.a. Nothing will happen to figure 1 as both the sides have 30 N.
2.b. The force with 30 N will push 10 N because 10 N is less force than 30 N.
6 0
3 years ago
A working model used to test a design is called a
Lisa [10]

Answer:

A working model used to test a design is called a

stop

5 0
3 years ago
When the ph rises from 10 to 12 , how many more times has the solution become basic?
koban [17]
2 times

Hope this helps


3 0
3 years ago
Complete orbital diagrams (boxes with arrows in them) to represent the electron configuration of valence electrons of carbon bef
kramer

Answer:

Explanation:

Before Hybridization => C: 1s²<u> ⇵</u> 2s²<u>⇵</u> 2p₋₁<u>↑</u> 2p₀<u>↑</u> 2p₊₁<u>∅</u>

After Hybridization => C: 1s² 2sp¹ 2sp¹ 2p₀¹2p₊₁¹ => 2 hybrid orbitals and 2 unhybridized p-orbitals at the n=2 energy level.

3 0
3 years ago
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