Chemistry, 03.07.2020 22:01, gokusandjimp6blzh

If 45.5 g of a metal that has a density of 3.65 g/mL is placed in 45.0 mL of water, what is the final volume?

Answers: 2

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Answers

answered: clarissajimenez27

The final volume of the metal and water is 54.46mL

Explanation:

Hello,

To solve this question, we'll first of all find the volume of the metal and assuming there's no loss of water by overflow in the container, we'll add the volume of the metal to the volume of the water to get the final volume.

Data;

Mass of the metal = 45.5g

Volume of the water = 45mL

Density of the metal (ρ) = 3.65g/mL

Density of the metal = mass / volume

ρ = mass / volume

Volume (v) = mass / density

Volume = 45.5 / 3.65

Volume = 12.46mL

The volume of the metal is 12.46mL.

When the metal is added to the container 45mL of water, assuming no water was lost by overflow in the container, the final volume =

Final volume = volume of metal + volume of water

Final volume = 12.46 + 45.0

Final volume = 57.46mL

The final volume of the metal and water is 57.46mL

answered: Guest

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.