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Artemon [7]
3 years ago
13

An organism that is able to form nutritional organic substances from simple inorganic

Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

An autotroph is an organism that is able to form nutritional organic substances from simple inorganic substances such as carbon dioxide.

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During ______ the composition of a substance does not change
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During the Physical changes
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How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.
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One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
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Which alternative energy source would be expected to be the least dependent upon the weather?
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Of the three sources listed, geothermal energy 
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Which one of the following groups of chemical compounds is composed entirely of organic compounds?
guajiro [1.7K]

Answer:

D

Explanation:

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Option A is wrong because calcium phosphate is not an organic compound, although all the other compounds in the option are organic compounds.

Option B is wrong because calcium chloride is not an organic compound

Option C is wrong also as calcium sulphate is not an organic compound

Option D is correct as all are organic compounds

7 0
3 years ago
An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had
Vlad [161]

Answer : The time passed in years is 2.74\times 10^2\text{ years}

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}

t=274.64\text{ years}=2.74\times 10^2\text{ years}

Therefore, the time passed in years is 2.74\times 10^2\text{ years}

4 0
4 years ago
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