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kati45 [8]
3 years ago
7

Which of the following happens to the particles when the temperature of liquid water drops?

Chemistry
1 answer:
Vikki [24]3 years ago
6 0

Answer:

The Kinetic energy decreases

Explanation:

because the speed of the molecules slows down.

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24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!
erastovalidia [21]
I'd say b, but i'm not 100 percent sure.<span />
6 0
3 years ago
Read 2 more answers
What is the chemical equation( formula ) of the reaction of photosynthesis
LekaFEV [45]

Answer:

The process of photosynthesis is commonly written as: 6CO2 + 6H2O → C6H12O6 + 6O2.

Explanation:

This means that the reactants, six carbon dioxide molecules and six water molecules, are converted by light energy captured by chlorophyll (implied by the arrow) into a sugar molecule and six oxygen molecules, the products.

4 0
3 years ago
What are the half-reactions for a galvanic cell with Zn and Mg electrodes?
Alona [7]

the half-reactions

cathode : Zn²⁺ (aq) + 2e⁻ ---> Zn (s)  

anode : Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Zn²⁺(aq) + Mg(s)→ Zn(s) + Mg²⁺ (aq)

<h3 /><h3>Further explanation</h3>

Given

Zn and Mg electrodes

Required

The half-reactions for a galvanic cell

Solution

To determine the reaction of a voltaic cell, we must determine the metal that serves as the anode and the metal that serves as the cathode.

To determine this, we can either know from the standard potential value of the cell or use the voltaic series

1. voltaic series

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

So the metal on the left will easily undergo oxidation and function as anode

Since Mg is located to the left of Zn, then Mg functions as anode and Zn as a cathode

2. Standard potentials cell of Mg and Zn metals :

Mg2+ + 2e– → Mg E° = -2,35 V

Zn2+ + 2e– → Zn E° = -0,78 V

The anode has a smaller E°, then Mg is the anode and Zn is the cathode.

7 0
3 years ago
How many fluorine atoms are present in 125.0g of phosphorus pentafluoride?
vladimir2022 [97]

molar mass of PF5 = 125.966 g/mol

125 g PF5 × (1 mol PF5/125.966 g PF5) = 0.992 mol PF5

0.992 mol PF5 × (6.022 × 10^23 molecules PF5);

= 5.97 × 10^23 molecules PF5

Since there 5 fluorine atoms per molecule of PF5,

(5.97.× 10^23 molecules PF5) × (5 atoms F/1 molecule PF5)

= 2.99 × 10^24 atoms F

8 0
3 years ago
The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th
Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically  Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their  standard states.

  Compound                 ΔfHº (kJmol⁻¹)

        SO₂                             -296.8

         O₂                                    0

         SO₃                            -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) =  1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction  will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

                         = -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº +  CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

                     = -99,787.8 J x 1 kJ/1000 J  = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0
3 years ago
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