All categories

3 years ago

What happens to the density of a substance when it is heated? Explain your answer.

1 answer:

7 0

Heating a substance causes molecules to speed up and spread slightly further apart, occupying a larger volume that results in a decrease in density. Cooling a substance causes molecules to slow down and get slightly closer together, occupying a smaller volume that results in an increase in density.

From: www.middleschoolchemistry.com

You might be interested in

How many grams are there in 7.4 x 1023 molecules of H2SO4?

bazaltina [42]

7.4x10^23 = molecules of silver nitrate sample

6.022x10^23 number of molecules per mole (Avogadro's number)

Divide molecules of AgNO3 by # of molecules per mol

7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)

(I leave off the x10^23 because they both will divide out)

Use your periodic table to find the molar weight of silver nitrate.

107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3

Now multiply your moles of AgNO3 with your molar weight of AgNO3

1.229mol x 169.868g/mol = 208.767g AgNO3

4 0

3 years ago

A student mixed a small amount of iron filings and sulphur powder in a dish. He could not

Temka [501]

Answer:

$\huge \boxed{\mathrm{Carbon \ disulphide}}$

Explanation:

Carbon disulphide is the liquid that can be used to separate iron fillings and sulphur powder.

When carbon disulphide is poured into the dish, the sulphur powder gets easily dissolved in the carbon disulfide. The iron fillings are left to settle on the bottom of the dish.

The iron fillings can get seperated through filtration. When the mixture of sulphur powder and carbon disulphide gets completely evaporated, the sulphur powder is left over.

7 0

2 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>