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Alex777 [14]
4 years ago
15

Solutions

Chemistry
2 answers:
jek_recluse [69]4 years ago
8 0

Answer:

Its B

Explanation:

Varvara68 [4.7K]4 years ago
7 0
What it is made of?
A) Water and Salt- salt is solute, water is solvent
B) Water and Sugar- sugar is solute, water is solvent
C) Water and HCl- HCl is solute, water is solvent
D) Water and NaOH- NaOH is solute, water is solvent

Which of these substances would be a poor conductor of electricity?
Water and Sugar make a poor conductor because sugar does not ionize in water. The other 3 solutions have solutes that ionize and make good conductors.
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1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Jenny wants to test the electrical conductivity of two substances dissolved in water. She is preparing the containers for the ex
katen-ka-za [31]

Answer:

Volume of the solutions

This is the most important factor for her to control.

3 0
3 years ago
Non-examples of air pressure
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No pressure







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3 years ago
What is mandrake tuber
Rudiy27
It's botanical name would be Mandragora officianarum, and it has a humanoid shape. Its a murderous plant that grows from blood that grows in Mediterranean.
4 0
4 years ago
Write the full ionic equation and net ionic equation for sodium dihydrogen phosphate + calcium carbonate, sodium oxilate + calcl
My name is Ann [436]

Answer:

<em>Sodium dihydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<u>Net ionic equation</u>

2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<em>Sodium oxalate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)

<u>Net ionic equation</u>

C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)

<em>Sodium hydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)

<u>Net ionic equation</u>

HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)

Explanation:

Let's consider two kind of equations:

  • Full ionic equation: includes all ions and species that do not dissociate in water.
  • Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
4 0
3 years ago
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