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AURORKA [14]
2 years ago
14

Carbon fixation occurs during the light reactions.

Chemistry
2 answers:
Ksju [112]2 years ago
5 0

Answer:

True

Explanation:

Y_Kistochka [10]2 years ago
5 0

Answer:true

Good luck and God bless! :) :)

Explanation:

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Which of these molecules is nonpolar?<br> O A. PF 3 <br>O B. 02 <br>O C. CO <br>O D. CHCI​
zloy xaker [14]
B is the answer


please mark me brainlist ^^
5 0
3 years ago
The mass of an atom may be found by adding the
Darina [25.2K]

Answer:

protons and neutrons

Explanation:

5 0
2 years ago
The moon is 250,000 miles away. How many inches is it from earth?
Lapatulllka [165]

Answer:

1.584e10 = 15,840,000,000,000

Explanation:

250,000 miles

multiply the length value by 63360

25e4 x 63360

= 1.584e10

3 0
3 years ago
Read 2 more answers
A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

M.F=O_{2} H_{2}

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

\frac{34.00}{17.008}= 1.999 round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

3 0
3 years ago
Determine the number of moles present in 32.5 g aluminum chloride.
Artyom0805 [142]
The number of moles in 32.5g of aluminum chloride is approximately 0.250 moles.
3 0
3 years ago
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