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3 years ago

At what time of the day is most photosynthesis happens

1 answer:

8 0

I assume that it would be whenever the sun is at it's highest point because it is normally the hottest and covers the most surface area in a place. I'd say like 2pm maybe.

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at 30°C 80 g of sodium nitrate is dissolved in 100 g of water is a solution saturated and unsaturated or supersaturated

soldi70 [24.7K]

Answer:

Explanation:

thats the answer

4 0

3 years ago

How many moles of AgCl2 are in 5.78x1024 molecules of AgCl2?

Zolol [24]

Answer:

9.6 mol AgCl2

Explanation:

You have to use Avogadro's number: 6.023 x 10^23

5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2

3 0

3 years ago

A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your

yarga [219]

Answer:

1,081.1 units of heparin should be given to 298 lb person.

Explanation:

We are given:

Weight of the person = 298 lb = 135.143 kg

Conversion factor used:

1 lb = 0.4535 kg

Number of unit of heparin to be given in an hour = 8.0 units/kg

Number of units given to the patient weighing 135.143 kg :

$8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units$

1,081.1 units of heparin should be given to 298 lb person.

4 0

2 years ago

A 34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of

natka813 [3]

Answer:

The answer to your question is CH₄

Explanation:

Process

1.- Write the possible chemical reaction

CxHy + O₂ ⇒ CO₂ + H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

44 g of CO₂ ------------------- 12 g of carbon

93.5 g of CO₂ ---------------- x

x = (93.5 x 12) / 44

x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

18 g of water ------------- 2 g of hydrogen

76.5 g of water -------- x g of hydrogen

x = (76.5 x 2) / 18

x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

12g of carbon ------------- 1 mol

25.5g of carbon ---------- x

x = (25.5 x 1) / 12

x = 2.12 moles of carbon

1 g of hydrogen ----------- 1 mol

8.5 g of hydrogen -------- x

x = (8.5 x 1) / 1

x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

CH₄

7 0

2 years ago

Read 2 more answers

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

2 years ago