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Debora [2.8K]
3 years ago
13

PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
iogann1982 [59]3 years ago
8 0

2. is point source

3. is non point source

storchak [24]3 years ago
4 0
QUESTION 2: POINT SOURCE
QUESTION 3:NONE POINT SOURCE
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Each of the following reactions is allowed to reach equilibrium in a sealed container. For which of the reactions could you shif
natta225 [31]

Answer:

None of the reaction will be favored to the right by a decrease in pressure.

Explanation:

       CH₄(g) + 2O₂(g) ⇄ CO₂(g) + 2H₂O(l)  

       CaCO₃(s) ⇄  CaO(s) + CO₂(g)  

       Br₂(g) + 3Cl₂(g) ⇄  2BrCl₃(g)  

       2H₂S(g) + 3O₂(g) ⇄  2SO₂(g) + 2H₂O(g)

From Le Chatellier's principle, we must understand that pressure changes only affects reactions in gaseous phases. The second reaction will not be affected by pressure.

We are now left with three equations.

Also, increase in pressure favors sides with lower volume. We can know the volume from the coefficients in the equation. Now let us check the volumes:

          CH₄(g) + 2O₂(g) ⇄ CO₂(g) + 2H₂O(l)  

         3 moles of gases      3 moles of gases

        Br₂(g) + 3Cl₂(g) ⇄  2BrCl₃(g)  

         4 moles of gases     2 moles of gases

       2H₂S(g) + 3O₂(g) ⇄  2SO₂(g) + 2H₂O(g)

        5 moles of gases      4 moles of gases

None of the reaction will be favored to the right by a decrease in pressure.

The first reaction will not be affected by any change in pressure because the total number of moles on the two sides are equal.

The last two reactions will be favored to the right by increasing pressure and a decrease in pressure will favor the backward left reaction.

7 0
3 years ago
Read 2 more answers
The concentration of glucose inside a cell is 0.12 mM. Outside the cell the concentration of glucose is 12.9 mM. Calculate the c
Alenkasestr [34]

Answer:

The value of he change in Gibbs free energy ΔG = - 18.083 KJ

Explanation:

Given data

The concentration of glucose inside a cell is (P) = 0.12 m M

The concentration of glucose outside a cell is (R) = 12.9 m M

No. of  moles = 1.5 moles

The change in Gibbs free energy

ΔG = RT ㏑\frac{P}{R}

ΔG = 8.314 × 310 ㏑\frac{0.12}{12.9}

ΔG = - 12.055 \frac{J}{mole}

Since No. of  moles = 1.5 moles

Therefore

ΔG = - 12.055 × 1.5

ΔG = - 18.083 KJ

This the value of he change in Gibbs free energy.

7 0
3 years ago
The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2). 2NI3 Right a
yuradex [85]

The moles of I₂ will form from the decomposition of 3.58g of NI₃ is 0.0136 moles.

<h3>How we calculate moles?</h3>

Moles of any substance will be calculated as:

n = W/M, where

W = required mass

M = molar mass

Given chemical reaction is:

2NI₃ → N₂ + 3I₂

Moles of 3.58g of NI₃ will be calculated as:

n = 3.58g / 394. 71 g/mol = 0.009 moles

From the stoichiometry of the solution, it is clear that:

2 moles of NI₃ = produce 3 moles of I₂

0.009 moles of NI₃ = produce 3/2×0.009=0.0136 moles of I₂

Hence, option (3) is correct i.e. 0.0136 moles.

To know more about moles, visit the below link:

brainly.com/question/15303663

8 0
2 years ago
This gas law works only if temperature is recorded in ———— units
Vsevolod [243]

Answer:

Kelvin

Explanation:

  • Gas laws re the laws that relates properties of gases such as pressure, temperature and volume at different conditions.
  • Gas laws includes, Boyle's law, Charles law, grahams law and pressure law among others.
  • Temperature is one of the quantity of gases explained by gas laws. For gas laws to work, temperature should always be in Kelvin. Therefore, when given temperature in degrees, it is converted to Kelvin by adding 273, that is °C + 273 = K.
8 0
3 years ago
Dust mites are microscopic bugs that are too small to see with the dissecting microscope. to see the details of a dust mite's sk
aleksklad [387]
You are right it is D
5 0
3 years ago
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