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guajiro [1.7K]
4 years ago
13

What is the coefficient for calcium oxide? CaO(s) + CO2(g) → CaCO3(s)

Chemistry
1 answer:
VARVARA [1.3K]4 years ago
4 0

Answer:

The coefficient is 1

Explanation:

CaO(s) + CO2(g) -> CaCO3(s)

In the balanced equation, the coefficient for CaO is 1

The coefficient represents the number of moles of a compound in the stoichiometry of the reaction

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2. A pure substance has constant physical and chemical properties, while mixtures have varying physical and chemical properties (i.e., boiling point and melting point).

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The visible part of the electromagnetic spectrum consists of the colors that we see in a rainbow. Different colors correspond to
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Given the following rate equation, complete the sentences to identify the reaction order for each reagent, the overall reaction
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Answer:

1. First

2. Third

3. Fourth

4.remain the same as

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Given the reaction equation;

Rate= k[A] [B]^3

We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.

If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.

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3 years ago
When reacting calcium carbonate with hydrochloric acid, what action should be taken?
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6 0
4 years ago
A. Match each type of titration to its pH at the equivalence point.
maks197457 [2]

Answer:answers are in the explanation

Explanation:

(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.

pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.

(b). Equation of reaction;

HBr + KOH ---------> KBr + H2O

One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O

Calculating the mmol, we have;

mmol KOH = 28.0 ml × 0.50 M

mmol KOH= 14 mmol

mmol of HBr= 56 ml × 0.25M

mmol of HBr= 14 mmol

Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.

The pH here is greater than 7

(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL

= 0.10 M

Ka=Kw/kb

10^-14/ 1.8× 10^-5

Ka= 5.56 ×10^-10

Therefore, ka= x^2 / 0.20

5.56e-10 = x^2/0.20

x= (0.20 × 5.56e-10)^2

x= 1.05 × 10^-5

pH = -log [H+]

pH= - log[1.05 × 10^-5]

pH = 4.98

Acidic(less than 7)

(c). 0.5 × 20/40

= 0.25 M

Ka= Kw/kb

kb= 10^-14/1.8× 10^-5

Kb = 5.56×10^-10

x= (5.56×10^-10 × 0.5)^2

x= 1.667×10^-5 M

pH will be basic

3 0
3 years ago
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