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maksim [4K]
3 years ago
14

Can someone PLEASE help me with this im struggling bad

Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer:

  1. 4.5+2.34= 6.84
  2. 4.5-5 =-0.5
  3. 6.00+3.411= 9.411
  4. 3.4×2.32 = 7.888
  5. 7.77/2.3= 3.37

7. 1200×23.4=28080

9. = 78.512

10. =341.199

11= 7.45

12 =65.0023

13.=3400210.34

Explanation:

please mark me as branliest

hope it is helpful

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How many degrees of freedom does each of the following systems have? (Answer as a number, i.e., 1, 2, 3, etc.)
Len [333]

Answer:

Explanation:

Hello, since the Gibbs' phase rule states the following equation:

F=C-P+2

Whereas C is the number of components and P the present phases, you answers are:

1. F=1-2+2=1.

2. F=2-2+2=2.

3. F=2-2+2=2.

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4 years ago
Alyssa is trying to balance the following equation: Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20 What number should be used to replace th
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Answer:

2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20

Explanation:

Since there's no "?" shown in the equation, let's balance it and solve it entirely.

Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20

1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.

To do this, let's start by looking at the elements that are only presnet once on each side:

On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H  → to balance it, you need 4 water molecules on the products side.

6 0
3 years ago
The value of Ksp for PbCl2 is 1.6 ×10-5. What is the lowest concentration of Cl-(aq) that would be needed to begin precipitation
Dovator [93]

Answer:

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

Explanation:

Concentration of lead nitarte = [Pb(NO_3)_2]=0.010 M

Pb(NO_3)_2(aq)\rightleftharpoons Pb^{2+}(aq)+2NO_{3}^{-}(aq0

1 Mole of lead nirate gives 1 mole of lead ion.

Concentration of lead ion in the solution = 1\times 0.010 M= 0.010 M

Pb(Cl)_2(aq)\rightleftharpoons  Pb^{2+}(aq)+2Cl^{-}(aq0

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The value of K_{sp} for [tex]PbCl_2= 1.6\times 10^{-5}

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1.6\times 10^{-5}=0.010 M\times [Cl^{-}]^2

[Cl^-]=0.04 M

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

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4 years ago
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ValentinkaMS [17]

Answer:

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Answer:

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Explanation:

4 0
3 years ago
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