All categories

3 years ago

ASAPPP. Describe how you could determine the specific heat of a sample of a solid substance. You may

Chemistry

1 answer:

Allushta [10]3 years ago

8 0

Answer:

I would use calorimetric to determine the specific heat and I would measure the mass of a sample

Explanation:

I would use calorimetry to determine the specific heat.

I would measure the mass of a sample of the substance.

I would heat the substance to a known temperature.

I would place the heated substance into a coffee-cup calorimeter containing a known mass of water with a known initial temperature.

I would wait for the temperature to equilibrate, then calculate temperature change.

I would use the temperature change of water to determine the amount of energy absorbed.

I would use the amount of energy lost by substance, mass, and temperature change to calculate specific heat.

You might be interested in

Is this statement true or false?<br><br> Gymnosperms reproduce using seeds, but angiosperms do not.

spayn [35]

Answer:

FALSE!

Explanation:

Gymnosperms do use seeds but are exposed like the pine cones of pines. Angiosperms still have seeds, however, they flower or fruit.

3 0

3 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

Calculate the molar solubility of AgCl, using its Ksp value of 1.8 x 10-10.

qwelly [4]

Answer:1.34 ×10^-5

Explanation:

AgCl = (Ag+) (Cl-)

=(S)(S)

= S²

S stands for molar solubility

Ksp =S² = 1.80 × 10^ -10

S= √[1.80 × 10^ -10]

=1.34 × 10^-5

8 0

3 years ago

What physical properties can be used to separate heterogeneous mixture's?

marysya [2.9K]

Evaporation, distillation, filtration and chromatography
Hope it help you, have a nice day

7 0

3 years ago

The breaking down or wearing away of rock and soil is known as... A. earthquake

Nataly_w [17]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers