What was the original volume of a gas if heating it from 22k to 85k produced a volume of 3.8 liters?

Drag the terms to the appropriate blanks to complete the following sentences summarizing the redox reactions of photosynthesis.

quester [9]

<h3>Answer;</h3>

1. In the light reactions, light energy is used to oxidize H2O to O2.

2. The electrons derived from this oxidation reaction in the light reactions are used to reduce NADP+ to NADPH.

3. The Calvin cycle oxidizes the light-reactions product NADPH to NADP+.

4. The electrons derived from this oxidation reaction in the Calvin cycle are used to reduce CO2 to G3P.

<h3>Explanation;</h3>

In the light reactions, light energy is used to remove electrons from (oxidize) water, producing O2 gas. These electrons are ultimately used to reduce NADP+ to NADPH.
In the Calvin cycle, NADPH is oxidized back to NADP+ (which returns to the light reactions). The electrons released by the oxidation of NADPH are used to reduce three molecules of CO2 to sugar (G3P), which then exits the Calvin cycle.
As ATP and NADPH are used in the Clavin cycle, they produce ADP and NADP+, respectively, which are returned to the light reactions so that more ATP and NADPH can be formed.

5 0

1 year ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

1 year ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0

1 year ago

The volume of blood plasma in adults is 3.1 L. it's density is 1.03 g/cm3. Approximately how many pounds of blood plasma are the

e-lub [12.9K]

The important thing in this question is the unit. The mass equals density * volume. 3.1 L = 3.1 * 10^3 cm3. So the mass is 3.193*10^3 g. 1 pound = 453.95 g. So the answer is 7.04 pounds.

3 0

2 years ago

Read 2 more answers

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

algol [13]

Answer:

820 K

Explanation:

As per Boltzman equation, kinetic energy (KE) is in direct relation to the temperature, measured in absolute scale Kelvin.

KE α T.

Then, the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be 3 times such temperature.

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

Second, convert 0°C to kelvin:

T (K) = T(°C) + 273.15 K= 273.15 K

Then, 3 times gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

7 0

1 year ago