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3 years ago

Which type of reaction occurs when 50mL quantities of 1 M Ba(OH)2 (aq) and H2SO4 (aq) are combined

Chemistry

1 answer:

iragen [17]3 years ago

5 0

Double replacement.

Ba(OH)2 + H2SO4 ====> 2H2O + BaSO4

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Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3

taurus [48]

1 $cm ^{3}$ = 0.001 $dm ^{3}$ . Therefore 200 $cm ^{3}$ = 0.2 $dm ^{3}$ . Molarity = $\frac{number of moles of NaCl}{volume of the solution} = \frac{0.1}{0.2} = 0.5 mol/dm^{3}$ . Hope this helps.

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3 years ago

A 155 g sample of an unknown substance was heated from 25°c to 40°c. in the process, the substance absorbed 569 calories of ener

Tasya [4]

Energy= 2381 joules
heat= Mass(kg) *change in temperature(K) * Cp
2381=0.155*(15)*Cp
Cp=1024 J/kg K

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3 years ago

Read 2 more answers

Balance this equation Hgo —> Hg + O2

Nesterboy [21]

Answer:

2Hgo —>2Hg + O2

Explanation:

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3 years ago

How many liters of oxygen gas are there in 45.0g O, at 0 degrees C and 1 atm of pressure? make sure to round to the correct numb

ZanzabumX [31]

Answer:

You can use the pdf to help you

Explanation:

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3 0

2 years ago

Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Reil [10]

Answer:

The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is 394.91 K.

Explanation:

The formula for molal boiling Point elevation is :

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b}$ = elevation in boiling Point

$K_{b}$ = Boiling point constant( ebullioscopic constant)

m = molality of the solution

i = Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

$Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}$

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

i = Van't Hoff Factor = 5

m = 8.5 m

$K_{b}$ = 0.512 °C/m

Insert the values and calculate temperature change:

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b} = 5\times 0.512\times 8.5$

$\Delta T_{b} = 21.76 K$

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

$\Delta T_{b} = T_{b} - T_{b}_{pure}$

$T_{b}_{pure}$ = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0

3 years ago