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3 years ago

Increasing the temperature gives particles more

2 answers:

7 0

Increasing the temperature a reaction takes place at increases the rate of reaction. At higher temperatures, particles can collide more often and with more energy, which makes the reaction take place more quickly

Arada [10]3 years ago

3 0

Answer:

i believe the answer is kinetic energy

Explanation:

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Consider the following reaction at 298K.

dmitriy555 [2]

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

<h3>Eªcell > 0; n = 2</h3>

8 0

3 years ago

What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 5

charle [14.2K]

Answer:

The volume we need is 8.89 mL

Explanation:

We analyse data:

36 % by mass → 36 g of HCl in 100 g of solution

Solution's density = 1.179 g/mL

5.20L → Volume of diluted density

pH = 1.70 → [H⁺] = 10⁻¹'⁷⁰ = 0.0199 M

HCl → H⁺ + Cl⁻

0.0199

pH can gives the information of protons concentrations, so as ratio is 1:1, 0.0199 M is also the molar concentration of HCl

Let's verify the molar concentration of the concentrated solution:

We convert the mass to moles: 36 g / 36.45 g/mol = 0.987 moles

As the solution mass is 100 g, we apply density to find out the volume:

Density = Mass / volume → Volume = Mass / Density

Volume = 100 g / 1.179 g/mL → 84.8 mL

Let's convert the volume from mL to L in order to define molarity

84.8 mL . 1L/ 1000mL = 0.0848 L

Molarity → 0.987 mol / 0.0848L = 11.6M

Let's apply the dilution formula:

M concentrated . V concentrated = M diluted . V diluted

11.6 M . V concentrated = 0.0199M . 5.20L

V concentrated = (0.0199M . 5.20L) / 11.6M → 8.89×10⁻³L

We can say, that the volume we need is 8.89 mL

8 0

3 years ago

Could anyone answer this please.?

igomit [66]

Answer:

I believe the answer is C.

Explanation:

epic gamer explanation: C + 02 = CO2

5 0

3 years ago

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ladessa [460]

Answer:

The molarity of the solution is 1.1 $\frac{moles}{liter}$

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

In this case

number of moles of solute= 0.564 moles
volume= 0.510 L

Replacing:

$molarity=\frac{0.564 moles}{0.510 L}$

Solving:

molarity= 1.1 $\frac{moles}{liter}$

The molarity of the solution is 1.1 $\frac{moles}{liter}$

4 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