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3 years ago

How do producers and consumers get energy from the sun?

1 answer:

6 0

Answer: In photosynthesis, producers combine carbon dioxide, water, and sunlight to produce oxygen and sugar (their food). Other organisms get energy by eating producers. ... It cannot directly use the Sun's energy to make food. As a consumer, it has to eat— or, consume— other organisms for energy.

Explanation: Thats how both producers and consumers get energy

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Which geologic action may produce a chain of volcanic islands?

aleksley [76]

Answer:

a converging plates

Explanation:

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3 years ago

If you have nitrate and carbon oxygen what can that combine with to make

WARRIOR [948]

Answer: Gunpowder

Explanation:

Gunpowder is made by mixing potassium nitrate with two fuels (atoms that can combine with oxygen atoms and release energy), carbon (charcoal) and sulfur. The oxygen atoms leave the nitrates and move onto the carbon and sulfur atoms, releasing a buttload of energy.

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3 years ago

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You have placed a can of lemonade in a cooler of is ice.what is the best way to describe what is happening to the ice

Firdavs [7]

<span>Heat from the can is making the water molecule move faster, which melts the ice.</span>

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4 years ago

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How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

zloy xaker [14]

Answer:

$m_{HCl}=36.1gHCl$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

$n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2$

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

$m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl$

Regards!

4 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago