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4 years ago

What chemical reaction shows photosynthesis

Chemistry

1 answer:

notka56 [123]4 years ago

6 0

Answer:

Chemical equation for photosynthesis

A water + carbon dioxide + sunlight → oxygen + glucose. B carbon + oxygen → carbon dioxide. C hydrogen + oxygen → water.

Explanation: This same photosynthesis cycle is usually expressed as: 6CO2 + 6H2O + 6O2 + C6H12O6. This implies that the solute, six carbon dioxide molecules plus 6 hydroxyl groups, are translated into something like a sugar molecule or six oxygen molecules, the goods, exploiting light energy absorbed the chlorophyll (implied by arrow).

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calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

Using the Hoffman apparatus for electrolysis, a chemist decomposes 2.3 moles of water into its gaseous elements. How many grams

solong [7]

Answer:

2.318032g

Explanation:

-The electrolysis equation of water is written as below:

$2H_2O_{(l)}->2H_2_{(g)}+O_2_{(g)}$

-The mole ratio of Water to the hydrogen formed is 1:1, therefore 2.3 moles of hydrogen gas is produced.

-Hydrogen's molar mass is 1.00784 grams:

$Mass=moles\times molar \ mass\\\\\=2.3\times 1.00784\\\\=2.318032\ g$

Hence, 2.318032 grams of hydrogen is produced.

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4 years ago

Which word describes the forces acting on the car?

KIM [24]

C. Unbalanced describes the forces acting on the car

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What is the affect of attraction or pull gravity between the sun and moon

REY [17]

I think the sun would suck in the moon

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3 years ago

25 points!!! please Which types of elements areconsidered neutraland Where are these types of elements found on the periodic tab

posledela

Elements which composed of an equal amount of three components protons, neutrons and electrons are considered neutral elements. example of neutral elements are hydrogen, helium, lithium and beryllium...

7 0

3 years ago