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q = mC∆T
q = (30.0g)(0.900J/goC)(50oC)
q = 1350 J
So, the right answer is 1350 J
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Answer:
false
Explanation:
first of all;-energy lead to an indotermic reaction.
indotermic is a reaction that absorbs energy \
*it has positive enthalpy of reaction
*Heat content of product is greater than that of reactant
*Heat is added to reactant side
example;- CO^2+2H^2+891kj --------- CH4 +2O2
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Answer: An atom with 6 protons, 5 electrons, and 7 neutrons
Explanation: In this case, neutrons do not matter as they have a charge of 0, or no charge. A proton has a charge of +1 and an electron has a charge of -1. Since there are 6 protons, the total charge of the protons would be +6. Since there are 5 electrons the total charge of the electrons would be -5. +6 - 5 would result in a charge of +1. This means that this atom would have an overall charge of + 1. Basically, if there is one more proton than electron, then the overall charge of the atom will be +1 but if there is one more electron than proton, then the overall charge of the atom will be -1.
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V