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Makovka662 [10]
3 years ago
7

Which do you think will cool down faster in? The teaspoon of water or the cup of water? Why?

Chemistry
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

I think the teaspoon will cool down faster.

Explanation:

The teaspoon of water will cool down faster because it is a smaller amount of water so the entire [drop] will absorb heat energy faster.

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Which drug presents the greatest challenges to forensic science
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Tobacco is the number one drug
5 0
3 years ago
Examine the given reaction. NH4NO3(s) → NH4+(aq) + NO3–(aq) ΔH° = 25.45 kJ/mol ΔS° = 108.7 J/mol·K Which of the given is correct
kobusy [5.1K]

Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

ΔG = ΔH - TΔS

Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.

Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

<h3>B)−6,942 J /mol</h3>

8 0
3 years ago
What is the correct way to represent 5600 L using scientific notation ??
Pachacha [2.7K]

the answer is 5.6 because you have to move the decimal to the left three times


5 0
3 years ago
Read 2 more answers
ASAP
Oksanka [162]

Answer:

Answer B

Explanation:

By definition mixtures are combinations of substances physically prepared and physically separated. Mixtures are classified as 'homogeneous mixtures' and 'heterogeneous mixtures'. For homogeneous mixtures all components are dissolved, with solute evenly distributed throughout the solution and assumes state of solvent. 'Heterogeneous mixtures' have inconsistent properties throughout and solute concentration is not consistent throughout the mixture.

6 0
3 years ago
Fe3O4(s) + 4H2(g)3Fe(s) + 4H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings w
alexandr402 [8]

Answer:

dS= 1.79*169.504

j/k = 303.41 j/k

Explanation:

Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)

dS(Fe3O4) =146.4 j/k

dS(H2) =130.684

dS(Fe) =27.78

dS(H2O) =188.825

dSrxn = dS[product]-dS[reactants]

= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]

= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k

This is the dS for 1mole Fe3O4

for 1.79 mols Fe3O4

dS= 1.79*169.504 j/k = 303.41 j/k

7 0
3 years ago
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