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3 years ago

Objects that are hotter will generally emit _____.

1 answer:

5 0

They emit b) more radiation compared to cooler objects.

This is because hotter objects have more kinetic energy (in order for the particles to vibrate- and so more infrared radiation circulates in its surroundings.)

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You have an evacuated container of fixed volume and known mass and introduce a known mass of a gas sample. measuring the pressur

balandron [24]

Answer is: the gas undergoes a chemical reaction that has fewer gas particles in products than in reactants. pressure is directly proportional to number of particles, so the pressure decreased.
Ideal gas law: PV = nRT. If amount of substance decreased, pressure also decreased.

4 0

4 years ago

What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?

rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

5 0

3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago

A student dissolves of 15 g aniline in of a solvent with a density of . The student notices that the volume of the solvent does

VikaD [51]

The given question is incomplete. The complete question is ;

A student dissolves of 15 g aniline in 200 ml of a solvent with a density of 1.05 g/ml. The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

Answer: The molarity is 0.81 M and molality is 0.82 m

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$