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3 years ago

How many grams of sodium are in 10g. of aodium sulfate

Chemistry

2 answers:

Sav [38]3 years ago

6 0

You wander will be 10,00 grams

Anarel [89]3 years ago

3 0

Answer:

10g

Explanation:

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The chemical equation shown represents photosynthesis. Carbon dioxide plus A plus light with a right-pointing arrow towards B pl

astra-53 [7]

Answer: <u><em>Option B; It traps light energy and converts it into chemical energy.</em></u>

<h2>Explanation: This substance is chlorophyll. It is a pigment present in leaves of all plants. It absorbs light energy and provides it to carry out the process of photosynthesis. Light energy is converted into chemical energy, in form of NADPH and ATP, which can be used by plants for photosynthesis.</h2><h2></h2><h2>This pigment is present only in plants, so option A is incorrect.</h2><h2></h2><h2>This pigment only absorbs and transfers energy to other molecules, and is not associated with carbon dioxide directly, so option C and D are also incorrect.</h2>

3 0

3 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Calculate the lattice energy of CuBr(s) using a Born–Haber cycle.

In-s [12.5K]

The 2nd one I believe
Answer- 2nd

7 0

3 years ago

Please help im really stressed and doing 2 tests at the same time

DanielleElmas [232]

The answer is A, good luck

7 0

3 years ago