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2 years ago

Balance this chemical equation. Choose "blank" for the box if no other coefficient is needed. Writing the symbol implies "1."

Chemistry

1 answer:

DENIUS [597]2 years ago

4 0

Answer: $3NH_4OH+AlCl_3\rightarrow Al(OH)_3+3NH_4Cl$

Explanation: A double displacement reaction is one in which exchange of ions take place.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$3NH_4OH+AlCl_3\rightarrow Al(OH)_3+3NH_4Cl$

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Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)

ivanzaharov [21]

Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 kJ/mol
Al(OH)3(s)= −1277.0 kJ/mol
 H2O(l)= −285.8 kJ/mol
AlCl3(s) =−705.6 kJ/mol

Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products

products- reactants:

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 kJ per mole at 25°C

6 0

2 years ago

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

2 years ago

What molecule(s) could diffuse across the plasma membrane? question 31 options: disaccharides fe2 tryptophan amino acid ch4

nevsk [136]

The molecule that could diffuse across the plasma membrane is methane (CH4).

<h3>What is diffusion?</h3>

Diffusion is the movement of fluids or substances from regions of high concentration toward regions of lower concentration.

The plasma membrane is the semipermeable membrane that surrounds the cytoplasm of a cell. The semipermeability means that it allows some molecules through but blocks other substances.

The semipermeable plasma membrane readily allows the passage of small hydrophobic and polar molecules.

Therefore, the molecule that could diffuse across the plasma membrane is methane (CH4).

Learn more about semipermeability at: brainly.com/question/1652796

#SPJ1

7 0

1 year ago

Find the percent composition of each element in KMnO4

VARVARA [1.3K]

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

6 0

3 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is (3) 3 × 12.4 hours

To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
n - number of half-lives
x - remained amount of the sample, in decimals
 $t_{1/2}$ - half-life length
t - total time elapsed.

First, we have to calculate x and n. x is remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams × 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
 $(1/2) ^{n} =x$
 $(0.5) ^{n} =0.125$
$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
 $t_{1/2} = 12.4$
 $t_{1/2} *n = t$
 $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope

7 0

3 years ago

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