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3 years ago

Question 2

Chemistry

1 answer:

azamat3 years ago

8 0

Answer:

Explanation:

For A, 2Al(OH)3 contains 6H on the left side of the equation but only has 3H on the right side of the equation, thus its not balanced

For B, there is 2(Al2), which is 4, on the left side of the equation but only 2Al on the right side of the equation, so it is not balanced.

For D, theres a total of 6 H on the left side of the equation as there is (OH)3 and 3H, but none of the right side, so its not balanced.

C is the correct answer as

on the left side of the equation:

2 Al(OH)3 has 2Al, 6O, and 6H

on the right side of the equation:

Al2O3 has 2Al and 3O

3 H20 has 3H and 6O

this makes a total of 2Al, 6H and 6O, which is equal to the number of each element in the left side, thus it is balanced and is the right answer.

In the future for these type of qns, you want to add up the number of each element on the left, then the ride, and see compare the numver of element they have. if the number of all the elements on both sides are the same, the equation is balanced.

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Help plz I rly. Need it

babymother [125]

Answer:

It should be D.

Explanation:

3 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

$deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol$

b) To calculate the constant we have the following expression:

$lnK=-\frac{deltaG_{rxn} }{RT}$

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

c) The equilibrium pressure of O₂ over M is:

$K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm$

3 0

3 years ago

A biology student set up an experiment to study 9 mice. On day 1, they were measured for weight and length. Then the mice were p

Sophie [7]

The independent variable would be the variable in the research that is being manipulated by the researcher. In this case, it would be temperature in the cage as it is what is being manipulated and changed in the research design. The dependent variable would be the variable that is being studied so, for this case, it would be the length and the weight of the mice. The constants are the factors that might affect the dependent variable but is held constant or the same by the researcher throughout the experiment. These are the size of the cage, amount of food and the exercise wheel. The flaw that the scientist would be studying the length of the mice since I don't think the temperature has any effect on it. And base from he results, the change in lengths are not conclusive.

4 0

3 years ago

Which of the following does not serve as a way to neutralize the charge in a body? Question 3 options: A) Bringing the charged b

Lelechka [254]

Answer: B. Adding more protons to a positively charged body until the number of protons matches the number of electrons

Explanation:

took test got it right

8 0

3 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

8 0

3 years ago