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2 years ago

How to prepare 1m of a solution

Chemistry

1 answer:

jenyasd209 [6]2 years ago

5 0

Answer:

To prepare a 1 M solution, slowly add 1 formula weight of compound to a clean 1-L volumetric flask half filled with distilled or deionized water. Allow the compound to dissolve completely, swirling the flask gently if necessary.

Explanation:

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Jamie rolls a 6-sided die 30 times and determines that the experimental probability of rolling a 2 is Start Fraction 1 over 15 E

anyanavicka [17]

Answer:

Conduct more trials

Explanation:

Theoretical Probability can be defined as what someone is expecting to happen

Experimental Probability on the other hand, is defined as what actually happens.

Probability is usually calculated in the same way for experimental probability and that of theoretical probability. You divide the total number of possible ways in which a particular outcome can happen, by the total number of outcomes itself.

In Experimental probability, the more times a probability is tried, it gets closer and even more closer to theoretical probability.

So, for the question, Jamie should improve the number of tries more, so as to get his experimental probability results to be closer to the theoretical probability result.

8 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis

weqwewe [10]

Answer:

$\Delta H_{rxn}=-2043.999kJ$

Explanation:

$\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]$

Where $n_{i}$ and $n_{j}$ are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

$\Delta H_{f}^{0}$ is standard heat of formation and $\Delta H_{rxn}^{0}$ is standard enthalpy change for reaction at $25^{0}\textrm{C}$

So, $\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

or, $\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]$

or, $\Delta H_{rxn}=-2043.999kJ$

4 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

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3 years ago

if you are asked to help the laboratory assistant in separating the mixture which method will you see?

FromTheMoon [43]

Answer:

use secondary data. the normal method to use

3 0

3 years ago

We are all familiar with the general principles of operation of an internal combustion engine: the combustion of fuel drives out

Alexus [3.1K]

Answer:

The work done by the system is 100 J

Explanation:

Given details

The cross sectional area of the of the container is A = 100.0 cm^2 = 0.01m²

The total distance pushed by the piston is d = 10 cm = 0.10m

The total external pressure by which piston pushed is P = 100 kPa

From above data, the following relation can be used to determine the change in volume of the container

∆V = A * d

∆V = 0.01 * 0.10 = 0.001 m³

By using the following relation, the work done by the system is calculated as;

Work done W = P * ∆V

W = 100 * 0.001 = 0.1 kJ = 100 J

The work done by the system is 100 J

7 0

3 years ago

How to prepare 1m of a solution​

How to prepare 1m of a solution