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Drupady [299]
4 years ago
9

Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

of so42− for 4 mol of cu2+.
Chemistry
1 answer:
ki77a [65]4 years ago
6 0
<span>

</span>Cu_4(OH)_6(SO_4)
<span>
You have OH-  conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is  4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = \frac{2}{3} \times 2.3 \times 10^{-6}

= 1.53  \times 10^{-6} &#10;

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Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (
poizon [28]

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

<h3>ΔHrxn = -635.14kJ/mol</h3>
7 0
3 years ago
N2+3H2—&gt; 2NH3 What volume of NH3 at STP is produced if 25.0g of N2 is reacted with an excess of H2 ?
Butoxors [25]

Answer:

The answer to your question is 40 L of NH₃

Explanation:

Data

Volume of NH₃ = x

mass of N₂ = 25 g

mass of H₂ = excess

Balanced chemical reaction

                     N₂  +  3H₂   ⇒  2NH₃

Process

1.- Find the molar mass of N₂ and NH₃

N₂ = 14 x 2 = 28g

2NH₃ = 2[ 14 + 3] = 34 g

2.- Write a proportion to solve this problem

                    28 g of N₂ --------------- 34 g of NH₃

                     25 g of N₂ -------------   x

                     x = (25 x 34)/28

                     x = 30.36 g of NH₃

3.- Calculate the volume of NH₃

                     17 g of NH₃ -------------- 22.4 L

                     30.36 g of NH₃ --------  x

                     x = (30.36 x 22.4) / 17

                     x = 40 L

6 0
3 years ago
The maximum level of lead allowed in drinking water is 15 mg/kg. What is this concentration in units of parts per million?
Alecsey [184]
15 ppm
1.5 × 10-2 ppm
1.5 × 104 ppm
3.1 ppm
1.5 × 10-2 ppm
6 0
3 years ago
How much heat is needed to raise a 0.30 g piece of aluminum from 30.C to 150C? The specific heat of aluminum 0.900j/gC
OLEGan [10]

Answer:

hi

Explanation:

8 0
3 years ago
*CHEMISTRY*
Alekssandra [29.7K]
It is No change!!!!!!!
8 0
3 years ago
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