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sattari [20]
1 year ago
11

Suppose you had used carbon tetrachloride, a liquid of density 2.20 g/mL, to determine the actual volume measured by your pipet.

What differences would there be in mass of the liquid measured and the actual volume of liquid measured for carbon tetrachloride compared to the same values for water?
Chemistry
1 answer:
VARVARA [1.3K]1 year ago
4 0

Answer:

Carbon tetrachloride would be 2.2 fold heavier than water

Explanation:

Carbon tetrachloride (2.20g/mL) is denser than water (1.00g/mL)

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The recommended daily intake of potassium (K) is 4.725 grams. Assuming that every raisin contains 3.677 milligrams of K, how man
Pani-rosa [81]
3.677mg =0,003677g \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 1 \ raisin\\ 4,725g \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x\\\\ x=\frac{4,725g*1}{ 0,003677g}\approx 1285 \ raisins
3 0
2 years ago
Consider the compounds 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. using silica gel plates, which of the two compoun
grandymaker [24]
1- we know that 4-tert-butylcyclohexanol is more polar than 4-tert-butylcyclohexanone (where the alcohols in general are more polar than ketons due to the hydrogen bond)
2- during separation via chromatography (in this case) the more polar solute will dissolve easily in polar solvents, where like dissolves like.
3- So, 4-tert-butylcyclohexanol will dissolve in ethyl acetete (which is polar) more than 4-tert-butylcyclohexanone, i.e, will have much higher Rf.
4- And also 4-tert-butylcyclohexanone will dissolve in dichloromethane (which lower in polarity than ethyl acetate) more than 4-tert-butylhexanol, i.e, will have much higher Rf 
3 0
2 years ago
Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin
sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0
2 years ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
2 years ago
Given the following equation and heat of reaction: 2 moles of A combine with 2 moles of B to produce 6 moles of C and 26 kJ What
Hoochie [10]

Answer: The change in enthalpy will be -13.

Explanation:-

Endothermic reactions are those in which heat is absorbed by the system and exothermic reactions are those in which heat is released by the system.

\Delta H for Endothermic reaction is positive and  \Delta H for Exothermic reaction is negative.

2A+2B\rightarrow 6C+26kJ  

2A+2B\rightarrow 6C       \DeltaH=-26kJ

When 1 mole of A combining with 1 mole of B to produce 3 moles of C

Thus as the stoichiometry has got half of the original , enthalpy of the reaction will also get half.

Thus for reaction :

A+B\rightarrow 3C+\frac{26}{2}kJ  

A+B\rightarrow 3C       \DeltaH=-13kJ

Thus the change in enthalpy will be -13.

8 0
2 years ago
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