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Alika [10]
3 years ago
6

The drop-counter is misaligned with the buret, causing it to miss counting every one out of five drops of base added to the HCl.

This error means the calculated molarity of the HCl solution will be ___________ its actual concentration.
Chemistry
1 answer:
Airida [17]3 years ago
4 0

Answer:

Less than

Explanation:

According to the question, the drop counter is counting four out of five drops of base added to the HCl. So when the titration will reach to the end point the volume of Base recorded will be less than the actual volume of the base dropped in the solution.

We will use law of equivalence for the calculation of concentration of HCl,

M_1V_1 = M_2V_2

Where, M_1 and V_1 are molarity and volume of HCl and M_2 and V_2 are molarity of volume of the base added.

Thus concentration of HCl will be calculated as,

M_1=\frac{M_2V_2}{V_1}

As the recorded V_2 is less than the actual amount present in the solution, the calculated value of concentration of HCl will also be less than the actual.

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A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel
postnew [5]

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

         \rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}

or,     n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

where,   n' = no. of formula units per cell

           A_{c} = molecular weight of cation = 90.5 g/mol

           A_{a} = molecular weight of anion = 37.3 g/mol

          V_{c} = volume of cubic cell = 3.55 g/cm^{3}

           a = edge length of unit cell = 3.9 \times 10^{-8} cm

        N_{A} = Avogadro's number = 6.023 \times 10^{23}

          \rho = density = 3.55 g/cm^{3}

Now, putting the given values into the above formula as follows.

           n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

                      = \frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}

                     = 0.9

                    = 1 (approx)

Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.

3 0
3 years ago
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 Ă— 10-13. Determine the molar sol
Kaylis [27]

Hey there!:

Given the reaction:

 NaBr  ⇌  Na⁺ +  Br⁻

  ↓              ↓         ↓

0.150M      0.150M     0.150M


AgBr ⇌ Ag⁺   + Br⁻

 ↓            ↓          ↓

 x            x         0.150M


Therefore:

Ksp = x * 0.150

x = ( 7.7 * 10⁻¹³ ) / 0.150

x = 5.1 * 10⁻¹²


Answer B


Hope that helps!

5 0
3 years ago
Which statement below best describes the relationship of the reaction below?
timurjin [86]
For every two AB produced, the reaction requires three A
3 0
3 years ago
Read 2 more answers
There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 ​ C, the volume is changed to 5.5 liters and the
HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0
4 years ago
A 1.00 x 10^-10 M solution of chloric acid Please help me solve this to find pH and pOH​
Svetradugi [14.3K]

Answer: The pH of solution is 10.

The pOH of the solution is 4.

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

As given concentration of acidic solution is 1.00 \times 10^{-10} M. Therefore, pH of the solution is calculated as follows.

pH = -log [H^{+}]\\= -log (1.00 \times 10^{-10})\\= 10

The relation between pH and pOH is as follows.

pH + pOH = 14

pOH = 14 - pH

= 14 - 10

= 4

Thus, we an conclude that pH of solution is 10 and pOH of the solution is 4.

3 0
4 years ago
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