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dezoksy [38]
3 years ago
11

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions.

It has the unwieldy formula of
KHC,H,O. This is often written in shorthand notation as KHP.
What volume of a 0.275 M calcium hydroxide solution is needed to exactly neutralize 2.48 grams of KHP?
ANSWER:



NO LINKS THAT IS ANNOYING
Chemistry
1 answer:
navik [9.2K]3 years ago
3 0

How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4?

  5.0 (328)

Ph. D. in organic chemistry with 6+ years of teaching experi..

Write down the balanced reaction first

3NaOH + H3PO4   makes    Na3PO4 + 3H2O

 

Given 0.10M NaOH (0.10mol/lit NaOH)

0.2M H3PO4 ( 0.2 mol/lit H3PO4)

15 ml of H3PO4 or 0.015 lit of H3PO4

Find volume of NaOH needed to neutralize the solution

Relate moles of H3PO4 with the moles of NaOH with the helps of coefficients in front of NaOH and H3PO4 seen in balanced equations.

               moles ↔ Litre use definition of molarity (moles/litrs)

Start with what is given for H3PO4

0.015 lit of H3PO4 x 0.2 moles of H3PO4 / lit of H3PO4 x 3 moles of NaOH/1 mole of H3PO4 x lit of NaOH/0.1 mol of NaOH = 0.09 lit or 90 ml.

The same colors in above equations cancel our leaving lit of NaOH as final answer.

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Answer:0.286M

Explanation:

10.7g / 149.89g/mol×0.25L

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What is the percent of H in NH3
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\frac{h3}{nh3}

\frac{1.008 \times 3}{(14.01) + (1.008 \times 3)}  \times 100\% =
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When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) What is the theoretical yield of Li
anyanavicka [17]

Answer:

The % yield of the reaction = 27.5 %

Explanation:

Step 1: Data given

Mass of Li = 12.7 grams

Mass of N2 = 34.7 grams

Actual yield of Li3N = 5.85 grams

Molar mass of  Lithium = 6.94 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of LI3N = 34.83 g/mol

Step 2: The balanced equation:

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles of Lithium

Moles Li = mass Li / Molar mass Li

Moles Li = 12.7 grams / 6.94 g/mol

Moles Li = 1.83 moles

Step 4: Calculate moles of N2

Moles N2 = 34.7 g/ 28 g/mol

Moles N2 = 1.24 moles

Step 5: Limiting reactant

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

Lithium is the limiting reactant. It will completely be consumed (1.83 moles).

N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles

There will remain 1.24 - 0.305 = 0.935 moles

Step 6: Calculate moles of Li3N

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N

Step 7: Calculate mass of Li3N

Mass Li3N =moles LI3N * Molar Mass LI3N

Mass Li3N = 0.610 moles * 34.83 g/mol

Mass Li3N = 21.2463 grams = Theoretical yield

Step 8: Calculate % yield

% yield = actual yield / theoretical yield

% yield = (5.85 / 21.2463)*100% = 27.5%

The % yield of the reaction = 27.5 %

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Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.

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