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3 years ago

Which of the following are acid formulas? (Choose 2)

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

HNO3 & H2SO4

Explanation:

HNO3 is Nitric acid and H2SO4 is Sulphuric acid

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What is the charge associated with each side of the HBr molecule?

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Answer:

Partial positive on hydrogen and partial negative charge on oxygen atom

Explanation:

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2 years ago

Day and night are caused by what

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Answer:

THE GOD ALMIGHTY!

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3 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

Answer: The mass of methanol that must be burned is 24.34 grams

Explanation:

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

4 0

3 years ago

A piece of fossilized wood has a carbon-14 radioactivity that is 1/4 that of new wood. the half-life of carbon-14 is 5730 years.

Marizza181 [45]

Answer:

1.146 x 10⁴ year.

Explanation:

The decay of carbon-14 is a first order reaction.

The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.

The integration law of a first order reaction is:

kt = ln [A₀]/[A]

k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.

t is the time = ??? years.

[A₀] is the initial percentage of carbon-14 = 100.0 %.

[A] is the remaining percentage of carbon-14 = 1/4[A₀] = 25.0 %.

∵ kt = ln [Ao]/[A]

∴ (1.21 x 10⁻⁴ year⁻¹)(t) = ln (100.0%)/[25.0 %]

(1.21 x 10⁻⁴ year⁻¹)(t) = 1.386.

∴ t = 1.386/ (1.21 x 10⁻⁴ year⁻¹) = 1.146 x 10⁴ year.

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3 years ago

What is the percent composition of each element in the compound below: CrPO4

Paha777 [63]

Chromium phosphate pentahydrate

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3 years ago