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Liula [17]
4 years ago
15

If you are to measure 55 uL of a liquid for an experiment, which pipette should you use and what value should be displayed on it

s window?
Select one:
a. p20: 550
b. p200: 055
c. p20: 055
d. p200: 550
e. p1000: 055

Chemistry
1 answer:
Alina [70]4 years ago
8 0

Answer:

\boxed{\text{b. P200; 055}}

Explanation:

1. Type of pipette  

The three most common types of micropipette are shown in the table below.  

\begin{array}{ll} \textbf{Type} &\textbf{V/ $\mu$L }\\ \text{P20} & 2 - 20 \\ \text{P200} & 20 - 200 \\ \text{P1000} & 200 - 1000 \\ \end{array}\\ \text{The only type that will dispense 55 µL is the $\boxed{\textbf{P200}}$}

2. Window  

The window on a P200 micropipette shows three digits.  

The first digit is hundreds, the second is tens, and the third is units of microlitres.  

A measurement of 55 µL will display as 055.  

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Mushrooms have gills?
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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
Which element requires the least i.e. potassium or gallium why
-Dominant- [34]

Answer:

Potassium

General Formulas and Concepts:

<u>Chem</u>

  • Reading a Periodic Table
  • Periodic Trends
  • Ionization Energy - energy required to remove an electron from a given element
  • Coulomb's Law
  • Shielding Effect
  • Z-effective and Forces of Attraction

Explanation:

The Periodic Trend for 1st Ionization Energy is increasing up and to the right. That means He would have the highest I.E and therefore take the most amount of energy to remove an electron.

Potassium and Gallium are both in Period 4. Potassium is element 19 and Gallium is element 31.

Potassium's electron configuration is [Ne] 4s¹ and Gallium's electron configurations is [Ne] 4s²3d¹⁰4p¹. Since both are in Period 4, they have the same number of core e⁻. Therefore, the shielding effect is the same.

However, since Gallium is element 31, it has 31 protons compared to Potassium, which is element 19 and has 19 protons. Gallium would have a greater Zeff than Potassium as it has more protons. Therefore, the FOA between the electrons and nucleus of Ga is much stronger than that of K. Thus, Ga requires <em>more</em> energy to overcome those FOA to remove the 4p¹ e⁻. Since K has less protons, it will have a smaller Zeff and thus less FOA between the e⁻ and nucleus, requiring <em>less</em> energy to remove the 4s¹ e⁻.

7 0
3 years ago
Benzene is 92.3% carbon and 7.7% hydrogen. In anexperiment,
Marysya12 [62]

Answer:

The molecular formula of benzene = C_{6}H_{6}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 92.3

Molar mass of C = 12.0107 g/mol

% moles of C = \frac{92.3}{12.0107} = 7.6848

% of H = 7.7

Molar mass of H = 1.00784 g/mol

% moles of H = \frac{7.7}{1.00784} = 7.6401

Taking the simplest ratio for C and H as:

7.6848 : 7.6401

= 1 : 1

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12+ 1 = 13 g/mol

Molar mass = 78.0 g/mol

So,  

Molecular mass = n × Empirical mass

78.0 = n × 13

⇒ n ≅ 6

<u>The molecular formula of benzene = C_{6}H_{6}</u>

7 0
4 years ago
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