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3 years ago

Approximately how much of the world’s oil and natural gas reserves are believed to be in the arctic?

1 answer:

7 0

In 2009, the US Geological Survey estimated the Arctic may be home to 30% of the planet's natural gas reserves and 13% of oil.

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When are kids ready to assume adult responsibilities?

Serga [27]

Answer:

18 years old or when they become mature.

Explanation:

5 0

3 years ago

you have a lead ball with a mass of 420 g. the density of lead is 10.5 g/cm³. what is the volume of the ball

IrinaK [193]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass = 420 g

density = 10.5 g/cm³

The volume is

$volume = \frac{420}{10.5} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

The process of generating electricity from dams and nuclear power plants is very different, but some of the energy transformatio

Sindrei [870]

Answer:

yuh

Explanation:

hi i just wanted to say i love ur pfp

3 0

3 years ago

Given the balanced equation 2C4H10 + 13O2 → 8CO2 + 10H2O, how many moles of CO2 are produced when 14.9g of O2 are used?

Anna [14]

Answer: The number of moles of $CO_2$ produced are, 0.287 moles.

Explanation : Given,

Mass of $O_2$ = 14.9 g

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{14.9g}{32g/mol}=0.466mol$

Now we have to calculate the moles of $CO_2$

The balanced chemical equation is:

$2C_4H_{10}+13O_2\rightarrow 10H_2O+8CO_2$

From the reaction, we conclude that

As, 13 mole of $O_2$ react to give 8 moles of $CO_2$

So, 0.466 mole of $O_2$ react to give $\frac{8}{13}\times 0.466=0.287$ mole of $CO_2$

Therefore, the number of moles of $CO_2$ produced are, 0.287 moles.

3 0

3 years ago

What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen

Blizzard [7]

Answer: $Mg_{1}Cl_{2}O_{8}$

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg= $\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles$

Moles of Cl = $\frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles$

Moles of O = $\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = $\frac {0.45}{0.45}=1$

For Cl = $\frac {0.89}{0.45}=2$

For O= $\frac {3.58}{0.45}=8$

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is $Mg_{1}Cl_{2}O_{8}$

4 0

3 years ago