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Nat2105 [25]
3 years ago
15

33.6 grams of CaO should have been produced in the decomposition of

Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

answer answer answer answer answer

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Which of the following is NOT a homogeneous mixture? *<br> koolaid<br> soda<br> tea<br> water
Wittaler [7]

Answer:

water

Explanation:

Because it does not have more than one component.

4 0
3 years ago
What is the volume of 1 lb of mercury. the density of mercury is 13.546g/ml?
joja [24]
<span>Take the inversion of density: 1mL/13.6 g and multiply it by the conversion factor 453.6 g/ 1 lb and the given 5.00 lb. Units for mass (grams) and units for weight (lbs) cancel leaving only units of volume. I believe it should be 167 mL or 0.167 L</span>
7 0
3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water
motikmotik

Answer:

0.153M

Explanation:

57.3/97.994 (molar mass)=0.585 moles of H3PO4

.0585/3.820L=0.153M

5 0
3 years ago
Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)
lesantik [10]

Answer:

N2(g) + 3H2(g) → 2 NH3(g)

Explanation:

N2(g) + H2(g) → NH3(g)

We start equaling the number of N atoms in both sides multiplying by 2 the NH3.

N2(g) + H2(g) → 2 NH3(g)

So we equals the H atoms (there are six in products sites)

N2(g) + 3 H2(g) → 2 NH3(g)

7 0
3 years ago
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