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vichka [17]
3 years ago
13

Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the steps in the correct order.

Chemistry
1 answer:
nexus9112 [7]3 years ago
5 0

This question is incomplete, the complete question is;

Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the following steps in the correct order;

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

Answer:

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.  

Explanation:  

First of all, We use a suction device to suck up liquid into pipet. Then we will fill the pipet up to the 10 mL marks and will discard the initial two volumes of liquid and will take a final third volume. We will replace the bulb with index finger and will drain it by placing the tip of pipet at the wall of beaker and drain the liquid.

Arranged in the following steps correctly;

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

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4 0
3 years ago
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zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect
MArishka [77]

Answer:

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data

First, the data are arranged in decreasing order, from highest to lowest:

3. 5.6

2. 5.1

8. 5.1

1. 4.9

6. 4.9

5. 4.7

7. 4.5

4. 4.3

Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3

Qexp = (|5.6 - 5.1|)/1.3 = 0.385

From tables, at 95% confidence level, for n = 8, Qcrit = 0.526

Since 0.385 < 0.526, the value for week 3 is accepted.

7 0
3 years ago
If 4520 kj of heat is needed to boil a sample of water, what is the mass of water
Cerrena [4.2K]

Answer:

1,085g of water

Explanation:

If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:

                                                  Q= m*C*Δt  where;

Q in the heat, in this case: 4520kj

m is the mas

Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.

In order to determine the mass, I will have the same equation but finding m

                                          m= Q/C*Δt    without   m=Q/C

So: m= 4,520J/4.18J/g°C

      m= 1,0813 g

5 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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