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2 years ago

Which type of energy is the tree changing the light energy into?

Chemistry

2 answers:

inysia [295]2 years ago

6 0

Answer:

chemical energy

In this case plants convert light energy into chemical energy, (in molecular bonds), through a process known as photosynthesis. Most of this energy is stored in compounds called carbohydrates. The plants convert a tiny amount of the light they receive into food energy.

Explanation:

jonny [76]2 years ago

4 0

Answer: Chemical energy

Explanation: In this case plants convert light energy (1) into chemical energy, (in molecular bonds), through a process known as photosynthesis. Most of this energy is stored in compounds called carbohydrates. The plants convert a tiny amount of the light they receive into food energy.

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1.25 moles of NOCl were placed in a 2.50 L reaction chamber at 427ºC. After equilibrium was reached, 1.10 moles of NOCl remained

kenny6666 [7]

Answer:

5.6 × 10^4

this is ur answer hope it helps u

5 0

3 years ago

A compound with molecular formula C6H9No that has an amide functional group and does not have an alkene group

Flauer [41]

Answer:

Following are the responses to the given question:

Explanation:

$\to C_6H_9NO$

calculating the HOI:

$= \frac{2C+2+N-H-X}{2} \\\\ =\frac{2\times 6+2+1-9-0}{2}\\\\ =\frac{12-6}{2}\\\\=\frac{6}{2}\\\\=3$

So, in the above compund there are three bounds or we can say that one is double and two bound is single.

Please find the attachment file of the Function group:

In the above given conditon compound doesn't include the alkene functional group that is absense of $-C=C-$ , and compound include $-C\equiv C-$ So, please find the attached file of the structure of the compound:

3 0

2 years ago

When 2.69 g 2.69 g of a nonelectrolyte solute is dissolved in water to make 345 mL 345 mL of solution at 26 °C, 26 °C, the solut

Gre4nikov [31]

Answer:

The molar concentration of this solution is 0.0463 mol/L

Explanation:

Step 1 : Data given

Mass of a nonelectrolyte solute = 2.69 grams

Volume of water = 345 mL = 0.345 L

Temperature = 26.0°CC = 273 + 26 = 299 K

The osmotic pressure = 863 torr

⇒ 863torr /760 = 1.13553 atm

Step 2: Calculate the molar concentration of this solution

Π = i*M*R*T

⇒with Π = the osmotic pressure = 1.13553 atm

⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1

⇒with M = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 299 K

1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K

M = 1.13553 / (0.08206*299)

M = 0.0463 mol/L

The molar concentration of this solution is 0.0463 mol/L

5 0

2 years ago

Given that a 12.00 g milk chocolate bar contains 8.000 g of sugar, calculate the percentage of sugar present in 12.00 g of milk

neonofarm [45]

So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!

5 0

3 years ago

Read 2 more answers

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago