Answers
Answer:
13440 J
Explanation:
c ≈ 4200 J / (kg * °C)
m = 80 g = 0,08 kg
= 10 °C
= 50 °C
The formula is: Q = c * m * ()
Calculating:
Q = 4200 * 0,08 * (50 - 10) = 13440 (J)
Related Questions
Assume a gasoline is isooctane, which has a density of 0.692 g/ml. What is the mass of 3.8 gal of the gasoline (1 gal = 3.78 l)?
What are two functions of the cilia?
Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solutionb. 72.8 g c2h6o in 2.34 l of solutionc. 12.87 mg ki in 112.4 ml of solution
Compare and contrast how observations and results can be used tosupport a conclusion to an experiment.
i. two
ii. three
iii. six
iv. eight
(b) How many atoms are directly bonded to the central atom in a trigonal bipyramidal molecule?
i. three
ii. four
iii. five
iv. six
(c) How many atoms are directly bonded to the central atom in an octahedral molecule?
i. three
ii. four
iii. six
iv. eight
Answers
Answer:
a) ii
b)iii
c)iii
Explanation:
three atoms directly bonded then only it is possible to achieve trigonal planar
trigonal bipyramidal means five atoms should attach to central atom
for octahedral six atoms must directly connected to central atom
Answers
Answer:
The reaction releases energy
Explanation:
The products of an exergonic reaction have a lower energy state (Delta-G) compared to the reactants. Therefore there is a negative delta –G between products and reactants after the reactions. This means some energy is lost into the environment usually through light or heat.
Final answer:
Exergonic reactions are characterized by a net release of energy but they still require a small initial energy input to start, referred to as the 'activation energy'. The speed or direction of the reaction is not determined by whether it's exergonic.
Explanation:
In the context of chemical reactions, the true statement for all exergonic reactions is that such reactions result in a net release of energy. However, even exergonic reactions, which are characterized by energy release, require a small initial input of energy to get started. This initial energy demand is referred to as the 'activation energy'. Also, it's important to note that the speed of the reaction or its directionality (whether it proceeds only in a forward direction) are not inherently determined by whether a reaction is exergonic. These aspects depend on other reaction conditions and catalysis.
Learn more about Exergonic Reactions here:
#SPJ3
Answers
I might be wrong, I’m just kinda going off based a little research and it says that it only happens in reptiles and teleost fish. The only option on there that’s related to it is the leatherback turtle.
Answers
Answer:
226.2 kJ/mol
Explanation:
Let's consider the following thermochemical equation for the combustion of acetylene.
C₂H₂(g) + (5/2) O₂(g) → 2 CO₂(g) + H₂O(l), ΔH°rxn = –1299 kJ/mol.
We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.
ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₂H₂(g)) - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × (-393.5 kJ/mol) + 1 mol × (-285.8 kJ/mol) - (-1299 kJ) - 1 mol × (0 kJ/mol)
ΔH°f(C₂H₂(g)) = 226.2 kJ/mol
Answer:
The enthalpy of formation of acetylene is 226.2 kJ/mol
Explanation:
Step 1: Data given
C2H2 (g) + (5/2)O2 (g) → 2CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Step 2: The balanced equation
The formation of acetylene is:
2C(s) + H2(g) → C2H2(g)
Step 3: Calculate the enthalpy of formation of acetylene
It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.
To have the reaction of the formation of acetylene we have to take:
⇒ the reverse equation of the combustion of acetylene
2CO2 (g) + H2O (l) → C2H2 (g) + (5/2)O2 (g)
⇒ The equation of formation of CO2 (multiplied by 2)
2C(s) + 2O2(g) → 2CO2(g)
⇒ the equation of formation of H2O
H2(g) + 1/2 O2(g) → H2O(l)
2CO2 (g) + H2O (l) + 2C(s) + 2O2(g) + H2(g) + 1/2 O2(g → C2H2 (g) + (5/2)O2 (g) + 2CO2(g) + H2O(l)
Final reaction = 2C(s) + H2(g) → C2H2(g)
Calculate the enthalpy of formation of acetylene =
ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol
ΔHf = 226.2 kJ/mol
The enthalpy of formation of acetylene is 226.2 kJ/mol
Answers
Answer: 22.8atm
Explanation:
T1 = —13°C = —13 +273 = 260K
T2 = 24°C = 24 +273 = 297K
V1 = V
V2 = 35% of V= 0.35V
P1 = 7atm
P2 =?
P1V1/T1 = P2V2 /T2
(7 x V)/260 = (P2 x 0.35V)/297
P2x0.35Vx260 = 7V x 297
P2 x 91V = 2079V
P2 = 2079V / 91V
P2 = 22.8atm
Answers
Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
The half-life of the breakdown reaction is 0.1066 h
The half-life of a substance is simply defined as the time taken for half of the original substance to decay.
The half-life of a first order reaction can be obtained by the following equation:
Where:
is the half-life
K is the decay constant
With the above formula, we can obtain the half-life of the breakdown reaction as follow:
Rate constant (K) = 6.5 h¯¹
Half-life (
) =.?
Therefore, the half-life of the breakdown reaction is 0.1066 h
Learn more: brainly.com/question/24286055?referrer=searchResults