All categories

2 years ago

Please help me answer this Will give extrasw brainslt

Chemistry

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

you can find the answers on safari

Explanation:

you can type the question but write nothing in the spots that dont have anything

You might be interested in

What is the difference between planets and in the solar system?

jenyasd209 [6]

Answer:

Some are large, some are small, some have more gravity then others. Some cant handle human life. Some are very cold some are very hot.

Explanation:

7 0

2 years ago

90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$

ΔᵣH° is negative, so the reaction is exothermic.

5 0

3 years ago

Ill give u brainliest pls help!

Pani-rosa [81]

radio waves, micro waves, infrared waves, visible light rays, ultra-violet rays, x-rays, then finally, gamma rays. I hope this was helpful to you.

4 0

3 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0

2 years ago

Bruh Im 4th on the leaderboard How is this possible

forsale [732]

Answer:

I guess you just answered a lot of questions

Explanation:

Thanks for the points btw :)

4 0

3 years ago