What molecules must the plant take in so the process of photosynthesis can occur

Please help! I have less than an hour left!

melisa1 [442]

Carbocation
I guess pls tell me if it wrong

8 0

3 years ago

You are given two unknown acids, HX and H2Y, and two values for Ka, 1x10^-7 and 1x10^9. You know that X is a much more electrone

Elena L [17]

Answer:

.

Explanation

In HX , X is more electronegative than Y so HX will ionise more because of ionic bond between H and X . On the other hand H₂Y will be less polar as compared to HX so it will ionise to a lesser extent . Hence Ka will be more for HX . Ka represents the degree of ionisation of acid . Higher the ionisation , higher is the value of Ka . H₂Y which is less polar will ionise less and hence it will have lesser value of Ka .

Hence H₂Y will have value of 10⁻⁷ and HX will have value of ka equal to 10⁹ .

6 0

4 years ago

A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod

sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

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7 0

1 year ago

A client who weighs 70 kg is receiving a solution of 0.9% sodium chloride (normal saline) 500 ml with dopamine 800 mg at 5 ml/ho

nordsb [41]

The mcg/kg/minute the client is receiving is 1.904 mcg/kg/min whose weight is 70 kg.

1 milligram (mg) = 1000 micrograms (mcg) or 0.001 grams (g) 1 g = 1000 mg 1 kilogram (kg) = 1000 g 1 kg = 2.2 pound (lb) 1 liter (L) = 1000 milliliters (mL)

The number of mcg/kg/minute = flow rate × concentration ÷ mass of client

Flow rate = 12 ml/hour = 12 ml/hour × 1 hr/60 min = 0.2 ml/min

Concentration = mass of dopamine/volume

where

mass of dopamine = 800 mg and

volume = 500 ml

Concentration =800 mg/500 ml

= 1.6 mg/ml

= 1.6 mg/ml × 1000 mcg/mg

= 1600 mcg/ml

mass of client = 70 kg

Calculating mcg/kg/minute

So, substituting the variables into the equation, we have

mcg/kg/minute = flow rate × concentration ÷ mass of client

mcg/kg/minute = 0.08 ml/min × 1600 mcg/ml ÷ 70 kg

mcg/kg/minute = 320 mcg/min ÷ 70 kg

mcg/kg/minute = 1.904 mcg/kg/min

Thus, the mcg/kg/minute the client is receiving is 1.904 mcg/kg/min

Learn more about mcg/kg/minute here:

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7 0

2 years ago

An acidic fog in pasadena was found to have a ph of 2.50. which expression represents this ph measurement?

joja [24]

Supposing a temperature of 25 degrees and supposing that all activity coefficients are 1

pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14

Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen ion, would be 10^(-2.50)

pH + pOH = 14
pOH = 14 - pH = 14 - 2.5 = 11.5
MOH- levels would be coordinated with pOH
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12

Therefore, MOH¯ = 3.2 × 10¯12 M

5 0

3 years ago