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Mashcka [7]
3 years ago
13

"In carrying out a titration of a hydrochloric acid solution with a standard sodium hydroxide solution, a student went beyond th

e end point before reading the volume on the burette. That is, the volume used was larger than the volume required to reach the end point. How will this error affect the calculated concentration of the hydrochloric acid?"
Chemistry
1 answer:
poizon [28]3 years ago
5 0

Answer:

The calculated concentration of HCl will be less than actual.

Explanation:

Suppose during titration, the <em>HCl</em> was taken in burette and the <em>NaOH</em> in the volumetric flask.

Now we will use equivalence formula for the calculation of concentration of HCl.

             N_{1} V_{1} = N_{2} V_{2}

Where L.H.S is for hydrochloric acid and R.H.S is for sodium hydroxide. The terms N and V represent normality and volume respectively.

If we calculate for

                              N_{1} = \frac{N_{2}V_{2} }{V_{1} }

We see that if the volume of the HCl is greater then the concentration of the HCl will be reduced.

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C. the rock may have diferrent textures 

7 0
3 years ago
Read 2 more answers
HCl gas is introduced at one end of the tube, and simultaneously NH3 gas is introduced at the other end. When the two gases diff
4vir4ik [10]

Answer:

When the two gases are mixed, the ammonium chloride precipitates in the tube walls.

Explanation:

This is the reaction:

HCl (g)  +  NH₃(g)  →  NH₄Cl (s) ↓

As the product formed is solid at room temperature, a suspension is first formed in the internal air of the tube that appears as a cloud. Afterwards it finally precipitates into the walls forming a white layer

5 0
3 years ago
Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O Wh
tamaranim1 [39]

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

4 0
3 years ago
16. Explain why antifreeze is used in vehicle radiators. Use your knowledge of the properties of
Fynjy0 [20]

Answer:

Antifreeze is whats used to keep your engine cool without freezing.

Explanation:

it keeps the engine from overheating.

It also prevents corrosion.

Here is a quote from google "Antifreeze works because the freezing and boiling points of liquids are “colligative” properties. This means they depend on the concentrations of “solutes,” or dissolved substances, in the solution. A pure solution freezes because the lower temperatures cause the molecules to slow down"

That quote is from "The Science Behind Antifreeze"

If you have any questions feel free to ask in the comments.

4 0
3 years ago
A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

<u>Step 2: Identify Conversion</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
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