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Alexeev081 [22]
2 years ago
15

What is the major product of the calvin-benson cycle that can then be used to form glucose?

Chemistry
1 answer:
noname [10]2 years ago
7 0
Glyceraldehyde 3 - phosphate
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1,5,25,125 what’s the pattern rule and extend by 3 more numbers
nika2105 [10]

For the first one the pattern is multiply the previous number by five as you see 1 x 5 = 5 and so on. To keep adding to it you would do

125 x 5 = 625     625 x 5 = 3125    3125 x 5 = 15625

Now for the second one the pattern is divide the previous number by three as you can see 2187 / 3 = 729 and so on. To keep going you would

81 / 3 = 27       27 / 3 = 9        9 / 3 = 3

I hope this helps you and if you have anymore questions i'll be  glad to answer them.



4 0
3 years ago
1. 17.0 grams of xenon hexafluoride is in a solid container. How many milliliters of that gas
BlackZzzverrR [31]

Answer: The volume of gas is 3020 ml

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 821.4 torr =  1.08 atm     (760 torr = 1atm)

V = Volume of gas in L = ?

n = number of moles = \frac{\text {given mass}}{\text {Molar mass}}=\frac{17.0g}{245.28g/mol}=0.069mol

R = gas constant =0.0821Latm/Kmol

T =temperature =302.7^0C=(302.7+273)K=575.7K

V=\frac{nRT}{P}

V=\frac{0.069mol\times 0.0821Latm/K mol\times 575.7K}{1.08atm}=3.02L=3020ml

Thus volume of gas is 3020 ml

4 0
2 years ago
Why do Group 12 elements have different properties than Group 13<br> elements?
blagie [28]
Group 12 Elements have two valence electrons while Group 13 Elements have three valence electrons.
Number of valence electrons tend to determine factors like reactivity. So elements with different number of valence electrons will have different properties.
That is why G12 and G13 have different properties
6 0
2 years ago
List two things in your environment that your body would need to adjust to
olchik [2.2K]

Answer:

Temperature & Humidity

Explanation:

If the body could not adapt to these changes, we wouldn't be able to live all over the planet; from the equator up to the polar caps.

6 0
2 years ago
Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t
Contact [7]

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)ΔH^{0}f(products) - ∑n(reactants)ΔH^{0}f(reactants)

where n = number of moles

ΔH^{0}f= enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

3 0
3 years ago
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