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barxatty [35]
3 years ago
10

If you expose a photosynthesizing plant to water that contains both radioactive h and radioactive o, in which of the products of

photosynthesis will the radioactive h and o show up? if you expose a photosynthesizing plant to water that contains both radioactive h and radioactive o, in which of the products of photosynthesis will the radioactive h and o show up? h in glucose and water; o in o2 h in glucose; o in water h in water; o in glucose h and o both in glucose
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0
I think the answer is H in glucose and water and O in O2. Photosynthesis is the process in which green plants and other organism convert light energy to chemical energy which is used in activities within the organism. Water that is exposed to the plant reacts with carbon dioxide to produce sugar and oxygen. The radioactive O will hence show up in the oxygen molecule and the radioactive H shows up in the sugar/ glucose molecule.
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Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w
quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0
2 years ago
What is an example
Paha777 [63]

Answer:

NaF sodium flouride

Explanation:

7 0
2 years ago
if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met
Bumek [7]

ANS: density = 13.41 g/ml

Density (d) of a substance is the mass (m) occupied by it in a given volume (v).

Density = mass/volume

i.e. d = m/v

m = (d) v -----(1)

The given equation from the plot of weight vs volume is :

y = 13.41 x ----(2)

Based on equations (1) and (2) we can deduce that the density of the metal is 13.41 g/ml

8 0
3 years ago
Read 2 more answers
Bryan and Alyssa are playing tug-of-war against Daniel and Maria. Bryan and Alyssa are exerting a combined force of 10 N. Daniel
mr_godi [17]

Answer:

D

Explanation:

4 0
3 years ago
PLEASE HELP
lisabon 2012 [21]

Answer:

The answer is B.

Explanation:

Trust me i took the test already its b.

6 0
3 years ago
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