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matrenka [14]
3 years ago
14

WILL MARK BRAINLIEST!! What is the percent composition of each element in barium nitrate?

Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0
<span>the percent composition of each element in barium nitrate is:</span><span>
52.56% Barium
10.71% nitrogen
36.73% oxygen

Hope this helps:)
</span>
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Consider molecular, complete ionic, and net ionic equations.
8090 [49]

Explanation:

(a) The molecular equations shows the equation in which all the species of the reactants and the products are in molecules and the net charge is zero.

The complete ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and are represented as ions.

The net ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and do not show the spectator ions which are same in the reactants and the products.

(b) If there is no spectator ions in the reaction, then the complete and the net ionic equations would be identical.

3 0
3 years ago
What are 3 sentences about the history of Egypt? Thank you for taking aside time to answer this. Please answer this as quickly a
Nataly_w [17]
1) The history of egypt is still unclear; artifacts, how the pyramids were built, gods and whatnot.
2) Egypt's history being uncovered in regards to curses and amulets is still being practiced in the country and some others because of the great influence it had.
3) Egyptians being able to construct such grand structures that still stand (pyramids) serves as evidence that proves Egypt was a very advanced civilization at the peak of their advancement.
5 0
3 years ago
Please help me due tomorrow
Dvinal [7]

  1. G
  2. A
  3. C
  4. I
  5. F
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Explanation:

this are answers

6 0
3 years ago
The molecular formula for butane is C4H10. Determine the percent composition for
masya89 [10]

%Mass

Ar C = 12 g/mol, Mr C₄H₁₀ = 58 g/mol, Ar H = 1 g/mol

\tt C=\dfrac{4.12}{58}\times 100\%=82.8\%\\\\H=100\%-82.8\%=17.2\%

or

\tt H=\dfrac{10.1}{58}\times 100\%= 17.2\%

4 0
3 years ago
Pure ethylene glycol, , is added to 2.00 kg of water in the cooling system of a car. The vapor pressure of the water in the syst
spayn [35]

Answer:

1367.7 g of ethylene glycol was added to the solution

Explanation:

In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')

525.8 mmHg - 451 mmHg = 451 mmHg . Xm

74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)

Xm = Mole fraction of solute / Moles of solute + Moles of solvent

We can determine the moles of solvent → 2000 g . 1  mol/18 g = 111.1 mol

(Notice we converted the 2kg of water to g)

0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent

0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute

18.4 moles = Moles of solute - 0.166 moles of solute

18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles

Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g

6 0
3 years ago
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