All categories

3 years ago

Which of the following showing the Law of Conservation of Mass?

Chemistry

2 answers:

Stells [14]3 years ago

7 0

The only one I know for sure is Mass is always conserved In a Chemical reaction.

sattari [20]3 years ago

6 0

Answer:

I'm pretty sure mass is always conserved during a chemical reaction.

Explanation:

Sorry for not answering the whole question.

You might be interested in

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<u> 394.91 K.</u>

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

<em>i =</em> 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

<em>i =</em> 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

2 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0

2 years ago

you are given an unknown gaseous binary compound. when 10.0g of the compound isburned in excess oxygen, 16.3g of water is produc

Flauer [41]

Since the compound has 1.38 time that of oxygen gas at the same conditions of temperature and pressure, we have the relationship:
MW/MWoxygen = 1.38
MW = 44.16

Since there is water formed during the reaction, the formula of the compound must be:
XaHb
where a and b are the coefficients of each element.

If the compound reactions with oxygen forming water and an oxide of the element X, the combustion reaction must be:
XaHb + ((2a + (b/2))/2) O2 = a (XO2) + (b/2)(H2O)
Using dimensional analysis:
10 (1/44.16) (b/2 / 1) (18) = 16.3
Solving for b:
b = 8

The compound now is XaH8. Most probably, the compound is C3H8 since it has a molecular formula of 44 and it reacts with O2 to form water and CO2.

5 0

3 years ago

Convert 3828 ml to L

crimeas [40]

Answer:

divide the volume value by 1000

So 3828/1000=3.828

3 0

3 years ago

Winter is coming and it's time to make sure you have the right amount of antifreeze

ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

molality

$\tt =\dfrac{107.608}{6}=17.93~m$

5 0

2 years ago