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How to study for a test on Isotopic Notations

poizon [28]

Just study it by looking it up

5 0

4 years ago

Which group of drugs includes legal painkillers like OxyContin?

Bumek [7]

It should be narcotics

4 0

4 years ago

Read 2 more answers

In a generic chemical reaction involving reactants A and B and products C and D, aA + bB → cC + dD, the standard enthalpy ΔH∘rxn

velikii [3]

Answer:

A. ΔH∘rxn = 21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

A.

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

ΔH∘rxn = 21.9 KJ/mol

B.

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

ΔH∘rxn = 103 KJ/mol

C.

Balanced equation for combustion of ethanol is:

C2H5OH + 3O2 → 2CO2 + 3H2O

3 0

3 years ago

50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0

creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA = 0.010 × 50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0

3 years ago

It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize w

Verdich [7]

The question is incomplete and complete question is :

It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.050 mL. Assuming the density of liquid water is 1.00 g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100°C and 1atm ?

Answer:

0.0028 moles of water are in one drop of liquid.

Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.

Explanation:

Volume of of drop = v = 0.050 mL

Mass of drop = m

Density of water = d = 1.00 g/mL

$m=d\times v=0.050 mL\times 1.00 g/mL=0.050 g$

Moles of water in drops:

$\frac{0.050 g}{18 g/mol}=0.0028 mol$

0.0028 moles of water are in one drop of liquid.

Pressure at which 0.0028 moles are vaporized = P = 1 atm

Temperature at which 0.0028 moles are vaporized = T = 100°C = 100+273 K = 373 K atm

Volume of moles of water = V

Moles of water = n = 0.0028 mol

$PV=nRT$ ( ideal gas equation )

$V=\frac{0.0028 mol\times 0.0821 atm L/mol K\times 373 K}{1 atm}$

V = 0.086 L = 86 mL ( 1L = 1000 mL)

Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.

6 0

3 years ago