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3 years ago

At 79.0°C, the volume of a gas is 49.0 L. If the pressure remains constant, at what

Chemistry

1 answer:

PtichkaEL [24]3 years ago

8 0

Answer:

At constant Pressure V /T = constant. sooooo

49 / 79 = 24.5 / T

T = 24.5 × 79/49 = 39.5°C

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What is an Atomic Model

NARA [144]

Answer:

the structure of an atom, theoretically consisting of a positively charged nucleus surrounded and neutralized by negatively charged electrons revolving in orbits at varying distances from the nucleus, the constitution of the nucleus and the arrangement of the electrons differing with various chemical elements.

Explanation:

8 0

3 years ago

Use Hess's Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given: 2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ

emmasim [6.3K]

Answer:

The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So, to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ

Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ

The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.

2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

2 ClF + 2 F₂ → 2 ClF₃

Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:

ClF + F₂ → ClF₃

Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

The enthalpy change (∆H) for the reaction is -108.7 kJ

3 0

3 years ago

REALLY NEED HELP. TAKING A TIMED TEST. 15 POINTS

zimovet [89]

The F atom . Oxygen has a -2 charge and fluorine has a +1 charge

7 0

2 years ago

The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp

EleoNora [17]

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

6 0

3 years ago

What is the heat of reaction (AH) of the following? C(s) + O2 (g) --> CO2 (g)

zlopas [31]

Answer:

C co2 2co enthalpy

2 Answers. Ernest Z. The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.

8 0

3 years ago