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fomenos
2 years ago
11

How many molecules of glucose are in 5. 72 grams of glucose c6h12o6?.

Chemistry
1 answer:
liq [111]2 years ago
4 0

Answer:

5.72 g ( 1 mol / 180.16 g ) (  6.022 x 10^23 molecules / mole ) = 1.90x10^23 molecules

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Plants use sunlight as energy to convert carbon dioxide and water into glucose and oxygen. Which best describes
masha68 [24]

Answer:

The law of conservation of mass indicates the same amount of carbon will be found in the reactants as in  the  products

Explanation:

The expression that best describes this reaction is that the same amount of carbon will be found in the reactants as in the products.

During the process of photosynthesis, carbon dioxide combines with water using solar energy to produce glucose and oxygen.

It can expressed as depicted below:

       6CO₂  +  6H₂O  →  C₆H₁₂O₆  + 6O₂

To validate our choice;

 This is a balanced chemical equation:

        6 moles of carbon is found on the reactant side and also, 6 mole of carbon on the product side.

But Glucose is not found on both sides of the expression

6 0
3 years ago
Read 2 more answers
Which material will heat the fastest?
Sholpan [36]

Answer:

Lead

Explanation:

Lower the specific heat faster it will heat, meaning liquid will heat up slowly than any other material mentioned above. Thus lead will heat up faster

8 0
3 years ago
What is the approximate mass of one mole of zinc?
VikaD [51]
The answer is (a) 30g. Zinc = 30. 1 mole = 30 x 1 = 30g
6 0
3 years ago
29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap
Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 161°C - 32°C

ΔT  = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c =  499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0
3 years ago
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0
3 years ago
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