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3 years ago

What is the environmental effect of 3NaCl

Chemistry

1 answer:

katrin [286]3 years ago

8 0

Answer:

Sodium and chloride ions separate when salts are dissolved in water. The dissolved sodium and chloride ions, in high concentrations, can displace other mineral nutrients in the soil. Plants then absorb the chlorine and sodium instead of needed plant nutrients such as potassium and phosphorus, leading to deficiencies.

Explanation:

the explanation is given inyour answer.

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Why do boys my age not like me ?

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2 years ago

Read 2 more answers

If a sample of HF gas at 694.9 mmHg has a volume of 3.463 Land the volume is changed to 5.887 L, then what will be the new press

VARVARA [1.3K]

Answer:

$\large \boxed{\text{381.7 mmHg}}$

Explanation:

Data:

p₁ = 694.9 mmHg; V₁ = 3.463 L

p₂ = ?; V₂ = 5.887 L

Calculation:

$\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{648.9 mmHg} \times \text{3.463 L} & = & p_{2} \times\text{5.887 L}\\\text{2247.1 mmHg} & = & 5.887p_{2}\\p_{2} & = & \dfrac{\text{2247.1 mmHg}}{5.887}\\\\& = &\textbf{381.7 mmHg}\\\end{array}\\\text{The new pressure of the gas is $\large \boxed{\textbf{381.7 mmHg}}$}$

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3 years ago

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.

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2 years ago

Compound that contains a terminal carbonyl?

malfutka [58]

Many compunds have a terminal carbonyl

Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.

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3 years ago