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3 years ago

7th grade science , help me please :))

Chemistry

1 answer:

konstantin123 [22]3 years ago

7 0

1. Organisms in just large molecules like carbohydrates proteins and fats and convert them into small molecules like carbon dioxide and water.
2. During cellular respiration of glucose molecule is gradually broken down into carbon dioxide and water. Then some ATP is produced directly in the reactance that transformers glucose.
3. Photosynthesis converts carbon dioxide and water into oxygen and glucose. While, cellular respiration converts oxygen and glucose into water and carbon dioxide.

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3. Calculate the amount of sodium carbonate required to neutralize 50 cc of N-H2SO4.

zlopas [31]

Answer:

234567890

Explanation:

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3 years ago

All of the facts about our star the Sun listed below are true EXCEPT which option?

Ronch [10]

Answer:

the sun is the smallest object in our Solar System

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3 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

3 years ago

Read 2 more answers

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago

What volume will be occupied by .756 mole of gas at 109 kPa and 30.5 degrees C

olga2289 [7]

Answer: it would be 0.026 moles

Explanation: PV=nRT, P is the pressure of gas, V is the volume it occupies n is the number of moles of gas present in the sample, R is the universal gas constant which is equal to 0.0821 atm L/mol K and T is the absolute temperature of the gas

8 0

3 years ago