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3 years ago

Where do plants produce food

Chemistry

2 answers:

marishachu [46]3 years ago

8 0

Photosynthesis,and stored food

spayn [35]3 years ago

7 0

Plants produce food in green leaves with CO2, H20, and sunlight. This process is call photosynthesis. Those with no leaves use their steams or roots. Hope this helps. :)

You might be interested in

The equation for another reaction used in industry isCO(g) + H₂O(g) <img src="https://tex.z-dn.net/?f=%5Crightleftharpoons" id="

Sloan [31]

Answer:

(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4

(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol

Explanation:

(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 3.2 mol 0 0 <em>Initial</em>

-x -x +x +x <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x 3.2-x x x <em>Equilibrium</em>

In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol

The amounts of CO and H₂O are:

CO = 2.0 - 1.6 = 0.4 mol

H₂O = 3.2 - 1.6 = 1.6 mol

The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:

Kc = (1.6x1.6)/(0.4x1.6)

Kc = 1.6/0.4

Kc = 4

(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 2.0 mol 0 0 <em>Initial</em>

-x -x +x +x <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x 2.0-x x x <em>Equilibrium</em>

Kc = (x*x)/((2.0-x)*(2.0-x))

4 = x²/(4 - 4x + x²)

16 - 16x + 4x² = x²

3x² - 16x + 16 = 0

Using Baskhara's equation:

Δ =(-16)² - 4x3x16

Δ = 256 - 192

Δ = 64

x = (-(-16) +/- √64)/(2*3)

x' = (16 + 8)/6 = 4

x'' = (16 - 8)/6 = 1.33

x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.

5 0

3 years ago

HELP PLZ!CHEMISTRY.WILL GIVE BRAINLIEST!!!!

Andrej [43]

1. 12 L = 12 dm³

2. 3.18 g

<h3>Further explanation</h3>

Given

1. Reaction

K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂

69 g K₂CO₃

2. 0.03 mol/L Na₂CO₃

Required

1. volume of CO₂

2. mass Na₂CO₃

Solution

1. mol K₂CO₃(MW=138 g/mol) :

= 69 : 138

= 0.5

mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5

Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :

= 0.5 x 24 L

= 12 L

2. M Na₂CO₃ = 0.03 M

Volume = 1 L

mol Na₂CO₃ :

= M x V

= 0.03 x 1

= 0.03 moles

Mass Na₂CO₃(MW=106 g/mol) :

= mol x MW

= 0.03 x 106

= 3.18 g

3 0

3 years ago

Read 2 more answers

Car A has a mass of 1.200 kg and is traveling at a rate of 22 km/hr. Car B has a mass of 1200 kg and is traveling in the

Zigmanuir [339]

Answer:

Car B has a greater momentum than car A

Explanation:

Given data:

Mass of car A = 1200 Kg

speed of car A = 22 km/hr

Mass of car B = 1200 Kg

speed of car B = 25 Km/hr

Solution:

For car A:

p = mv

p = 1200 Kg × 22 km/hr

p = 26400 kg.km/hr

For car B:

p = mv

p = 1200 Kg × 25 km/hr

p = 30,000 kg.km/hr

7 0

3 years ago

If a 1000. ml of linseed oil has a mass of 929 g, what is the density of the oil

uysha [10]

$Formula\ for\ density:\\\\
density=\frac{mass}{volume}\\\\
mass=929g=0,929kg\\\\
volume=1000ml=0,001m^3\\\\
denisty=\frac{0,929kg}{0,001m^3}=
\boxed{929\frac{kg}{m^3}}$

5 0

3 years ago

. In an experiment, 1.90 g of NH3 reacts with 4.96 g of O2. 4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(g) (i) Which is a limiting reactant

Simora [160]

1. The limiting reactant in the reaction is NH₃

2. The mass of the excess reactant remaining is 0.49 g

3. The mass of NO produced from the reaction is 3.35 g

<h3>Balanced equation </h3>

4NH₃ + 5O₂ —> 4NO + 6H₂O

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mass of NH₃ from the balanced equation = 4 × 17 = 68 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂ from the balanced = 5 × 32 = 160 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 4 × 30 = 120 g

SUMMARY

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂ to produce 120 g of NO

<h3>1. How to determine the limiting reactant </h3>

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂

Therefore,

1.90 g of NH₃ will react with = (1.90 × 160) / 68 = 4.47 g of O₂

From the calculation made above, we can see that only 4.47 g out of 4.96 g of O₂ given is needed to react completely with 1.90 g of NH₃.

Therefore, NH₃ is the limiting reactant.

<h3>2. How to determine the mass of the excess reactant remaining </h3>

Mass of excess reactant (O₂) given = 4.96 g
Mass of excess reactant (O₂) that reacted = 4.47 g

Mass of excess reactant (O₂) remaining =?

Mass of excess reactant (O₂) remaining = 4.96 – 4.47

Mass of excess reactant (O₂) remaining = 0.49 g

<h3>3. How to determine the mass of NO produced </h3>

In this case, the limiting reactant (NH₃) will be used.

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO

Therefore,

1.90 g of NH₃ will react to produce = (1.90 × 120) / 68 = 3.35 g of NO.

Thus, 3.35 g of NO were obtained from the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

5 0

2 years ago