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If 5.1 l of antifreeze solution (specific gravity = 0.80) is added to 3.8 l of water to make a 8.9 l mixture, what is the specif

Maslowich

Density of the mixture = mass of the mixture / volume of the mixture

Mass of the mixture = mass of antifreeze solution + mass of water.

Mass of antifreeze solution = density of the antifreeze solution * volume

Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g

Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g

Mass of mixture = 4080 g + 3800 g= 7880 g

Volume of mixture = volume of antifreeze solution + volume of water

Volume of mixture = 5100 ml + 3800 ml = 8900 ml

Density of mixture = 7800 g / 8900 ml = 0.876 g/ml

Specific gravity of the mixture = density of the mixture / density of water = 0.876

Answer: 0.876

3 0

3 years ago

How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?

statuscvo [17]

Answer:

A) $6.5\times 10^{23}\ \text{molecules}$

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

The reaction of electrolysis would be

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Number of moles of $H_2O$

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}$

From the reaction it can be seen that 2 moles of $H_2O$ gives 1 mole of $O_2$

So, number of moles of $O_2$ produced is

$\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}$

Number of molecules

$1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}$

So, $6.5\times 10^{23}\ \text{molecules}$ of oxygen is produced.

8 0

2 years ago

Which subatomic particle is involved in chemical bonding and is responsible for an element's reactivity?

Murrr4er [49]

There are three different subatomic particles present in the atoms of each element: neutron, proton and electron. It is the electrons, and more specifically the valence electrons, that determine the reactivity of an element.

6 0

3 years ago

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How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .

3 0

3 years ago

Which of the following best describes a mixture in which ionic compounds are dissociated in solution?

Ierofanga [76]

Ion solution is correct. hope it helps

6 0

3 years ago

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