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2 years ago

How many grams are in 4.3 x 10 -4 moles of calcium phosphate?

Chemistry

1 answer:

attashe74 [19]2 years ago

5 0

Explanation:

Ca3(PO4)2 1 mole has 310.2 g

Ca3(PO4)2 4.3 × 10^-4 moles has 4.3 × 10^-4 × 310.2

= 1,333.86 × 10^-4

= 1.334 × 10^-1

OR 0.1334 g

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How many molecules are there in 0.50 moles chlorine?

mariarad [96]

Answer:

molecules Cl- = 3.01 E23 molecules

Explanation:

molecules Cl- = 0.50 mol CL- * ( 6.02 E23 molecules/ mol ) = 3.01 E 23 molecules

3 0

3 years ago

Read 2 more answers

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0

3 years ago

If 200. g of water at 20°C absorbs 41 840 J of energy, what will its final temperature be? (Specific Heat of water is 4.184 J/g*

Elena-2011 [213]

Answer: The final temperature will be $70^0C$

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed =41840 J

c = specific heat = $4.184J/g^0C$

m = mass of water = 200 g

$T_{final}$ = final temperature =?

$T_{initial}$ = initial temperature = $20^0C$

Now put all the given values in the above formula, we get:

$41840J=200g\times 4.184J/g^0C\times (T_{final}-20)^0C$

$T_{final}=70^0C$

Thus the final temperature will be $70^0C$

3 0

3 years ago

Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro

attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the chemical reaction:

$NaN_3~->~Na~+~N_2$

The next step is to balance the reaction:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the mol calculation using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the molar ratio between $NaN_3$ and $N_2$ we can calculate the moles of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can calculate the grams:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

I hope it helps!

5 0

3 years ago

What needs to happen for a solute to dissolve in a particular solvent

sergiy2304 [10]

the atractions between the solute and solvent molecules must be greater than the atractions keeping the solute together and the atractions keeping the solvent togetherrr.

7 0

3 years ago