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3 years ago

Addition of a catalyst can speed up a reaction by providing an alternate reaction pathway that has a

Chemistry

1 answer:

kkurt [141]3 years ago

5 0

Answer:

Addition of a catalyst can speed up a reaction by providing an alternate reaction pathway that has a lower activation energy

Explanation:

A catalyst is an agent that increases the rate of a chemical reaction by providing an alternate pathway for the reaction that requires a lower activation energy. As the requirement for activation energy is less in the presence of a catalyst, there are more reactant particles becoming involved in the chemical reaction and as such there are more products formed per unit time, or there is an increase in the rate of the reaction

Example of catalyst include

1. Addition of potassium permanganate to hydrogen peroxide to aid in the rapid decomposition into water and oxygen

2. Platinum serves as a catalyst in the complete combustion of carbon monoxide into carbon dioxide.

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Determine the oxidation number of sulfur in S203 ( with steps)

sdas [7]

Answer:

The oxidation number of sulphur in is +2

Explanation:

S2O3 2-

2 (S) + 3(O)=-2

2 (S) + 3(-2)=-2

2 (S) + (-6) =-2

2 S = +4

S= +2

8 0

3 years ago

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

3 years ago

Blowing carbon dioxide gas into a solution of magnesium oxide will form what product?

AlekseyPX

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➷ It would form magnesium carbonate.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

3 0

3 years ago

Zinc phosphate is used as a dental cement. A 50.00-mg sample is broken down into its constituent elements and gives 16.58 mg oxy

cricket20 [7]

Answer:

Zn3P2O8

Explanation:

In this particular question, it is necessary to convert the respective masses to percentages. We convert to percentages by placing each mass over the total mass and multiplying by 100%. Since the total is 50mg, conversion to percentage can be done by multiplying the masses by 2 as 100/50 is 2

For Oxygen = 16.58 * 2 = 33.16%

For phosphorus = 8.02 * 2 = 16.04%

For zinc = 25.40 * 2 = 50.80%

We then proceed to divide these percentages by their respective atomic masses. The atomic mass of oxygen, phosphorus and zinc are 16, 31 and 65 respectively.

O = 33.16/16 = 2.0725

P = 16.04/31 = 0.5174

Zn = 50.80/65 = 0.7815

Now, we divide by the smallest value which is that of the phosphorus

O = 2.0725/0.5174 = 4

P = 0.5174/0.5174 = 1

Zn= 0.7815/0.5174 = 1.5

Now, we need to multiply through by 2. This yields: O = 8, P = 2 and Zn = 3

The empirical formula is thus: Zn3P2O8

7 0

3 years ago

? What is the oxidation number of B in HBO3? D-1 0 O +3 O +6

lesya692 [45]

H20* SOO N34 Thats my answer

6 0

3 years ago