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Roman55 [17]
4 years ago
5

76 PNTS PLZ HELP WILL GIVE BRAINLIST

Chemistry
1 answer:
melamori03 [73]4 years ago
3 0

The formula for calculating  the density (p)=m/v

m:mass

v:volume

<u>Explanation:</u>

In the question above,given that

<u>mass of water (m)</u> =100 g

<u>Volume of water in a beaker (v)</u>=100 mL

<u>The Density of the water(p</u>)=m/v=100 g/100 mL= 1 g/mL

The answer is (B) 1 g/mL

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Calculate the standard free-energy change at 25 ∘C for the following reaction:
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Answer:

Standard free-energy change at 25^{0}\textrm{C} is -3.80\times 10^{2}kJ/mol

Explanation:

Oxidation: Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)

Reduction: Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)

--------------------------------------------------------------------------------------

Overall: Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)

Standard cell potential, E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}

So, E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V

We know, standard free energy change at 25^{0}\textrm{C}(\Delta G^{0}): \Delta G^{0}=-nFE_{cell}^{0}

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, \Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol

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3 years ago
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Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
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Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

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