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2 years ago

a thermometer containing 8.3g of mercury has broken. if mercury ha a density of 13.6g/mL. what volume is spilled?

Chemistry

2 answers:

scoray [572]2 years ago

4 0

D = m / V

13.6 = 8.3 / V

V = 8.3 / 13.6

V = 0.610 mL

hope this helps!

enyata [817]2 years ago

3 0

Answer: The volume of mercury spilled is 0.61 ml.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of mercury= 8.3 grams

Density of mercury= $13.6g/ml$

Putting in the values we get:

$13.6g/ml=\frac{8.3g}{Volume}$

$Volume=0.61ml$

Thus the volume of mercury spilled is 0.61 ml.

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A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml. having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0

2 years ago

2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b

Virty [35]

Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: $2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl$

Compound Molar mass (g/mol)

NaOH 39.997

$MgCl_{2}$ 95.211

$Mg(OH)_{2}$ 58.3197

So, 2.50 g of NaOH = $\frac{2.50}{39.997}$ mol of NaOH = 0.0625 mol of NaOH

4.30 g of $MgCl_{2}$ = $\frac{4.30}{95.211}$ mol of $MgCl_{2}$ = 0.0452 mol of $MgCl_{2}$

According to balanced equation-

2 mol of NaOH produce 1 mol of $Mg(OH)_{2}$

So, 0.0625 mol of NaOH produce $(\frac{0.0625}{2})$ mol of NaOH or 0.03125 mol of NaOH

1 mol of $MgCl_{2}$ produces 1 mol of $Mg(OH)_{2}$

So, 0.0452 mol of $MgCl_{2}$ produce 0.0452 mol of $Mg(OH)_{2}$

As least number of moles of $Mg(OH)_{2}$ are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of $Mg(OH)_{2}$ would be produced = 0.03125 mol

= $(0.03125\times 58.3197)$ g

= 1.822 g

6 0

2 years ago

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?You do not need to look up any values to answer t

Nutka1998 [239]

Answer:

Reactions 1, 3 and 5

Explanation:

First thing's first, let's ensure that all the reactions given are balanced. This is given as;

CO(g) + 1/2 O2(g )→ CO2(g)

Li(s) + 1/2 F2(l) → LiF(s)

C(s) + O2(g) → CO2(g)

CaCO3(g) → CaO + CO2(g)

2Li(s) + F2(g) → 2LiF(s)

For the condition to be valid;

- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.

- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.

This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.

3 0

2 years ago

Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl

frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is $2.29\times 10^{4} gal$ .

Since, 1 gal= 3785.41 mL

Thus, $2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL$

Density of water is 1 g/mL thus, mass of water will be $8.66\times 10^{7}g$ .

Since, 1 grams of chlorine → $10^{6}$ grams of water.

1 g of water → $10^{-6}$ g of chlorine and,

$8.66\times 10^{7}g$ of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

$V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL$

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0

2 years ago

(b) When a sample of C2H5OH was combusted, the volume of CO2(g) produced was 18.0 L when measured at

myrzilka [38]

Answer:

i) The number of moles of CO₂ (g) produced from the reaction = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂ (g)

collected = 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction = 52.4 kJ

Explanation:

The balanced chemical reaction when ethanol is combusted is given as

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

The volume of CO₂(g) produced was 18.0 L when measured at 21.7°C and 1.03 atm.

i) Number of moles of CO₂ (g) produced by the reaction

With the correct and logical assumption that CO₂ is an ideal gas, the ideal gas equation has the relation

PV = nRT

P = pressure = 1.03 atm = 1.03 × 101325 Pa = 10,435.96 Pa

V = Volume of the gas = 18.0 L = 0.018 m³

n = number of moles = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 21.7 + 273.15 = 294.85 K

(10,435.96 × 0.018) = n × 8.314 × 294.85

n = 0.076629106 = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂(g)

collected.

Recall the stoichiometric balance of the reaction

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

2 moles of CO₂ is obtained from 1 mole of C₂H₅OH

0.07663 mole of CO₂ will be obtained from (0.07663×1/2) mole of C₂H₅OH; that is, 0.03831 mole of C₂H₅OH.

But we can convert this number of moles used up to mass of C₂H₅OH produced

Mass = (Number of moles) × (Molar Mass)

Molar mass of C₂H₅OH = 46.07 g/mol

Mass of C₂H₅OH combusted from the reaction

= 0.03831 × 46.07 = 1.765 g

But density of C₂H₅OH = 0.79 g/mL

Density = (Mass)/(Volume)

Volume = (Mass)/(Density) = (1.765/0.79)

= 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction.

The heat of combustion of C₂H₅OH at the temperature of the reaction = -1367.6 kJ/mol. (From literature)

1 mole of C₂H₅OH combusts to give 1367.6 kJ of heat

0.03831 mole of C₂H₅OH will give (0.03831×1367.6) = 52.39 kJ = 52.4 kJ

Hope this Helps!!!!

3 0

2 years ago