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Evgen [1.6K]
2 years ago
10

Why do plants link thousands of monosaccharides together to form polysaccharides such as starch?

Chemistry
1 answer:
Fittoniya [83]2 years ago
3 0

Answer: 2

Explanation:

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What is 25 milliliters in cubic centimeters
lubasha [3.4K]

1cc (cubic centimeter) and 1mL (milliliter) are the same volume. So, 25mL = 25cc

3 0
3 years ago
The structural formula for CF4 is
nevsk [136]

Answer:

PubChem CID 6393

Structure  

Find Similar Structures

Chemical Safety  

Compressed Gas

Laboratory Chemical Safety Summary (LCSS) Datasheet

Molecular Formula CF4

Molecular Weight  

88.004 g/mol

Explanation:

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3 years ago
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How many TOTAL carbon atoms are their in this equation? A. 1 B. 2 B.4 B. 6
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What equation, love?
8 0
3 years ago
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The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

7 0
3 years ago
Read 2 more answers
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
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