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Blababa [14]
3 years ago
5

A 30.0 mL sample of phosphoric acid is neutralized with 18.0 mL of a 3.00 M NaOH solution in a titration.

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer: The Answer is 0.3.

Explanation: Solved in the attached picture.

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Is NaCl CH3OH LiOH or H2SO4 a bronsted acid
vesna_86 [32]

Answer:

D. H₂SO₄

Explanation:

Bronsted acids are those that donate H+ ions. In this question, H₂SO₄ is a Bronsted acid.

Note: H₂SO₄ is one of seven strong acids that you should try to memorize.

3 0
3 years ago
Please help will mark you brainiest! Use word bank when answering the question!
zheka24 [161]

Answer:

Blank 1: varying

Blank 2: constant

Explanation:

7 0
3 years ago
Consider the reaction 2 S + 3 O2 → 2 SO3 , which has a 75.1% yield. How much O2 is consumed if 583 g of SO3 are produced?
Ymorist [56]

Answer:

A/1.      10.9 mol O2

Explanation:

583 g x 1 mol SO3 x 3 mol O2 /

      80.057 g mol SO3 x 2 mol SO3

- You just need to find molar mass of SO3, which is 80.057 g.

- Everything else came from formula. Further explanation...

- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.

- If you need a further explanation or more help on any problems I would be happy to help, just let me know.

4 0
3 years ago
How many electrons will a metal generally have in its outer energy level?
Zarrin [17]

Answer:

Metals have one or two electrons in their outermost shell

C. 1-2

Explanation:

  • Metals have low ionisation energy because they easily looses the outermost electrons
  • They have only one- two electrons in the outer most shell.
  • They loose these electron to form charged species called cation.

4 0
3 years ago
Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF
algol [13]

The molar concentration of the original HF  solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342

7 0
3 years ago
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