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3 years ago

Which one of the following is the net ionic equation for the reaction of nitric acid with aluminum hydroxide

1 answer:

5 0

The ionic eqn is as follow:

1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)

3moles of No3- ion on both sides cancels out to give the net ionic eqn:

1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)

You might be interested in

The 3 components of air

jok3333 [9.3K]

Nitrogen (around 78%), Oxygen (around 21%), and Argon (around 1%).

Hope this helps :)

8 0

2 years ago

Read 2 more answers

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

3 years ago

Which of the following will increase the number of collisions without changing the energy of reactant molecules?

natulia [17]

Answer:increasing the concentration of reactants

Explanation:

Collision is the phenomenon in which the reactant molecules come to nearest closness,as a result the reactants are converted into products.

Now the number of effective collision is directly proportional to the number of reactants added..

6 0

3 years ago

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At

anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

N2 3H2 2NH3

I [N2]₀ [H2]₀ 0

C - x -3x +2x

E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0

3 years ago

Read 2 more answers

You are required to prepare 500 ml of a 6.00 M solution of HNO3 from a stock solution of 12.0 M. Describe in detail how you woul

andriy [413]

Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

Explanation:

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = concentration of stock solution = 12.0 M

$V_1$ = volume of stock solution = ?

$C_2$ = concentration of diluted solution= 6.00 M

$V_2$ = volume of diluted acid solution = 500 ml

Putting in the values we get:

$12.0\times V_1=6.00\times 500$

$V_1=250ml$

Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

8 0

2 years ago