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2 years ago

The distance between Earth and its moon is 384000000 meters. Express this distance in kilometers.

1 answer:

3 0

384000000m

to change from m to km you just need to divide that give unit in Meters by 1000

which equals,
384000

so,

384000000 meters = 384000 kilometers

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How does a meteorologist use an understanding of what causes weather to predict a storm?

Svetllana [295]

Meteorologists also use satellites to observe cloud patterns around the world, and radar is used to measure precipitation. All of this data is then plugged into super computers, which use numerical forecast equations to create forecast models of the atmosphere.

8 0

2 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

2 years ago

an element in group two has two valence electrons. When forming a chemical bond it will most likely (gain, lose, share, or not r

Charra [1.4K]

Lose electrons - electrons want to fill their outer valence shell, so sometimes instead of gaining it is easier to lose some and have a filled outer shell

3 0

2 years ago

What mass of natural gas (CH4) must you burn to emit 264 kJ of heat?

mariarad [96]

The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.

Here 1 mol of CH4 realesed 802.3 KJ

To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4

The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol

The to obtain the mass multiply the number of moles times the molar mass:

mass = n * MM = 0.329mol * 16g/mol = 5.26 grams

Answer: 5.26 grams

7 0

2 years ago

A chemist mixes 96.2 g of chloroform with 31.2 g of acetone and 98.1 g of acetyl bromide. Calculate the percent by mass of each

stiv31 [10]

Answer:

42.6 % CHCl₃, 13.83 % C₃H₆O and 43.5 % CH₃COBr

Explanation:

This is a solution with 3 compounds:

96.2 g of CHCl₃

31.2 g of C₃H₆O

98.1 g of CH₃COBr

Total mass of solution is : 96.2g + 31.2 g + 98.1g = 225.5 g

Percent by mass is, the mass of the compounds in 100 g of solution

Let's prepare the rule of three

225.5 g of solution have ____ 96.2 g ___ 31.2 g _____98.1 g

100 g of solution would have ____

(100 . 96.2) / 225.5 = 42.6

(100 . 31.2) / 225.5 = 13.83

(100 . 98.1) / 225.5 = 43.5

5 0

2 years ago