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Annette [7]
3 years ago
13

In a titration, a few drops of an indicator are added to a flask containing 35.0 milliliters of HNO3(aq) of unknown concentratio

n. After 30.0 milliliters of 0.15 M NaOH(aq) solution is slowly added to the flask, the indicator changes color, showing the acid is neutralized.
What are the products of this reaction?
Chemistry
1 answer:
sasho [114]3 years ago
5 0

Answer:

NaOH(aq) + HNO3(aq)------>NaNO3(aq) + H2O(l)

Explanation:

A thing to note is that an acid and a base will react to form a metal salt + H2O.

~Hope it helps:).

You might be interested in
Bauxite must go through two processes to produce aluminum metal. The yield of the Bayer process, which extracts aluminum oxide f
Mashcka [7]

The yield of aluminium obtained from 1 m^3 of bauxite is 419810 g

<h3>What is a Percent yield</h3>

A percent yield of a substance measures the amount of the substance actually obtained as a percentage ratio of expected yield.

Percent yield = actual yield / expected yield × 100%

<h3>How to calculate the mass of aluminium obtained from bauxite </h3>

From the data given:

40 % of the bauxite is converted to aluminium oxide.

Volume of bauxite = 1 m^3

40 % of 1 m^3 = 0.4 m^3

volume of aluminium oxide = 0.4 m^3

density of aluminium oxide = 3965 kg/m^3

  • Using mass = density × volume

mass of aluminium oxide = 0.4 × 3965 kg

mass of aluminium oxide = 1586 kg

Formula of aluminium oxide is Al203

molar mass of aluminium oxide = 102 g

  • percentage mass of aluminium in one mole of aluminium oxide = mass of aluminium / mass of aluminium oxide × 100 %

Percentage mass of aluminium in aluminium oxide = 54/102 × 100

Percentage mass of aluminium in aluminium oxide = 52.94 %

Expected mass of aluminium from aluminium oxide = 52.94 × 1586

Expected mass of aluminium = 839.62 kg

Actual yield = 40 % × 839.62

Actual yield of aluminium = 419.81 kg

mass of aluminium in grams = 419810 g

Therefore, mass of aluminium obtained from 1 m^3 of bauxite is 419810 g

Learn more about percent yield at: brainly.com/question/8638404

3 0
3 years ago
Look at the potential energy diagram below. What amount of energy does the products have ?
12345 [234]
80kj my guy trust me
4 0
3 years ago
Read 2 more answers
How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?
babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps.  Let me know if anything is unclear.

8 0
3 years ago
According to the following pKa values listed for a set of acids, which would lead to the strongest conjugate base?
Ghella [55]
I think the answer is -2 which is answer A
6 0
3 years ago
Which scientist bombarded gold sheets with alpha particles to look at the atom?
Alekssandra [29.7K]

the correct scientist is rutherford


6 0
3 years ago
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