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3 years ago

What is a thunder’s volume

2 answers:

7 0

Lighting has the diameter of 1 to 2 inches. I'm sure it can get bigger or smaller, but it can also heat the air to 70,000 degrees Fahrenheit (39,000 degrees Celsius. It can do that in a few milliseconds, which is pretty darn fast.
Hope this information helped you out:)

katrin2010 [14]3 years ago

6 0

Lighting has the diameter of 1 to 2 inches. It can also heat the air to 70,000 degrees f

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Ch3chclch(ch3)ch2ch2ch2ch2br name the molecule iupac rules please

nadezda [96]

Answer:

1-bromo-6-chloro-5-methylheptane

Explanation:

There are 3 substituents, 1 chlorine, 1 bromine, and 1 methyl group.

The longest carbon chain consists of 7 carbons.

There are no double bonds, based on the saturation and substituents.

So you have a 1-bromo-6-chloro-5-methylheptane.

Hope that was the brainliest answer!

4 0

2 years ago

Which reagent is the limiting reactant when 0.400 mol Al(OH)3 and 0.400 mol H2SO4 are allowed to react

jeka94

Answer:

H₂SO₄ will be the limiting reagent.

Explanation:

The balanced reaction is:

2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction).

You can use a simple rule of three as follows: if by stoichiometry 2 moles of Al(OH)₃ reacts with 3 moles of H₂SO₄, how much moles of H₂SO₄ will be needed if 0.4 moles of Al(OH)₃ react?

$moles of H_{2} SO_{4} =\frac{0.4 moles of Al(OH)_{3}*3 moles of H_{2} SO_{4} }{2 moles of Al(OH)_{3}}$

moles of H₂SO₄= 0.6 moles

But 0.6 moles of H₂SO₄ are not available, 0.40 moles are available. Since you have less moles than you need to react with 0.4 moles of Al(OH)₃, H₂SO₄ will be the limiting reagent.

7 0

3 years ago

The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction

andreev551 [17]

Answer:

The NO + O3 is the dominant reaction.

Explanation:

First of all, let's convert to molecules/cm³;

For O3;

O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT

Thus, plugging in the relevant values to get;

n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)

So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =

9.84 x 10^(11) molecules/cm³

But from the question, NO has 2 moles, and thus concentration is;

2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³

For O2;

Following the same pattern for O3, we obtain;

(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³

Now, for NO and O3 reaction the rate is; k[NO] [O3]

Thus rate;

= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s

For 2NO + O2 → 2NO2 reaction, rate = k[NO]2 [O2]

Thus, rate;

= (2 x 10^(-38) cm^(6)/molec².s )( 1.968 x 10^(12) molec/cm³) ²

(5.167 x 1018 molec/cm³)

= 40,000 molec/cm³.s

Observing the two rates, it's clear that the NO + O3 is the dominant reaction.

6 0

3 years ago

Organize the following list of atoms from lowest to highest Electronegativity:

Anon25 [30]

Answer:

Barium<Strontium<Calcium <Magnesium< Beryllium

Explanation:

Electronegativity is defined as the ability of an atom in a bond to attract the shared electrons of the bond towards itself.

Electronegativity is a periodic trend that decreases down the group and increases across the period.

Hence, if i want to arrange Beryllium, Barium, Strontium, Magnesium, Calcium in order of increasing electronegativity, i will have;

Barium<Strontium<Calcium <Magnesium< Beryllium

5 0

3 years ago

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Ber [7]

Answer: C. The Number of Molecules

Explanation: I just took the test, it's correct.

6 0

4 years ago