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3 years ago

Plants carry out photosynthesis to make sugar from carbon dioxide and water. True or false?

Chemistry

2 answers:

lora16 [44]3 years ago

8 0

Answer:

la re spuesta en verdadero

Greeley [361]3 years ago

3 0

Answer:

True

Explanation:

During photosynthesis, plants take in carbon dioxide (CO2) and water (H2O) from the air and soil. Within the plant cell, the water is oxidized, meaning it loses electrons, while the carbon dioxide is reduced, meaning it gains electrons. This transforms the water into oxygen and the carbon dioxide into glucose.

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The average air tempature in an area over a long period of time is called

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3 years ago

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Why do you think the atomic number is basically represented by the number of protons and not the number of electrons?

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3 years ago

Gasoline and motor oil are chemically similar. They are both mixtures of nonpolar hydrocarbons

Softa [21]

Answer:

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3 years ago

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What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?

Vadim26 [7]

Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

Mass of methane CH4 (g) = 80.0 g.

Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

$PV=nRT.$

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

$80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.$

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

$V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}$

The volume would be 158 L.

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1 year ago