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3 years ago

What determines the severity of the storm?

Chemistry

2 answers:

Pepsi [2]3 years ago

6 0

Search up the answer bud

Mrac [35]3 years ago

5 0

Answer:

A thunderstorm is classified as “severe” when it contains one or more of the following: hail one inch or greater, winds gusting in excess of 50 knots (57.5 mph), or a tornado. ... There are about 100,000 thunderstorms each year in the U.S. alone. About 10% of these reach severe levels

Explanation:

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What would be the electron configuration of a sulfur ion S−2?

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Explain, in terms of particles, what happens when methane boils and why

Ronch [10]

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There are multiple butene isomers (Butene) and some (2-Butenes - cis and trans) actually have higher boiling point than n-Butane (there is also Isobutane, of course, with quite much lower boiling point than all of them) and some (1-Butene, Isobutylene) have lower, so this isn't really a fair or simple question. But on simplest level, it can again be said that 1-butene has lower boiling point because it has very similar shape but slightly lower molar mass (2H less) than n-butane.

Explanation:

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2 years ago

What is the atomic number of a sodium atom that has 11 protons and 12 neutrons?

galben [10]

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3 years ago

(I)how many atoms are present in 7g of lithium?

ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

Part (i) :

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

Part (ii) :

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

Part (iii) :

To calculate the moles of $F_2$ :

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

To calculate the moles of $CO_2$ :

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

To calculate the moles of $OH^-$ ions :

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

4 0

3 years ago