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SVEN [57.7K]
3 years ago
9

What is 93,000,000 in scientific notation?

Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0
I believe it is 9.3x10^7
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The second reaction in the formation of sulfuric acid occurs slowly.
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<u>Answer:</u> The correct option is NO_2(g)

<u>Explanation:</u>

A catalyst is defined as the chemical species that increases the reaction rate but does not participate in it and is left behind after the completion.

A homogeneous catalyst is one that is present in the same phase as the reactants and products.

A heterogeneous catalyst is one that is present in different phase as that of reactants and products.

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

As all the reactants and products are in gaseous state so, the homogeneous catalyst must also be in the gaseous state only.

Hence, the correct option is NO_2(g)

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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