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max2010maxim [7]
3 years ago
8

How many pounds in a 3.00 l bottle of drinking water?

Chemistry
1 answer:
slava [35]3 years ago
6 0

Answer:

             6.61 Pounds

Solution:

Step 1: Calculate Mass of Water as;

                        Density  =  Mass  ÷  Volume

Solving for Mass,

                        Mass  =  Density  ×  Volume   ------ (1)

As,

                        Density of Water  =  1 g.cm⁻³

And,

                        3 L of Water  =  3000 cm³

Putting values in equation 1,

                        Mass  =  1 g.cm⁻³  × 3000 cm³

                        Mass  =  3000 g

Step 2: Convert Grams into Pounds;

As,

                        1 Gram  =  0.002204 Pounds

So,

                        3000 Grams  =  X Pounds

Solving for X,

                      X =  (3000 Grams  ×  0.002204 Pounds)  ÷  1 Gram

                      X =  6.61 Pounds

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Sonja [21]
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8 0
3 years ago
The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c
olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

7 0
3 years ago
quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3
Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.  

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

Mᵣ:                 60.05            78.00

                3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g:           125               275  

2. Calculate the moles of each reactant

\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}

3. Calculate the moles of (CH₃COO)₃Al from each reactant

\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} =  \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol  (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol  (CH$_{3}$COO)$_{3}$Al}

\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}

5 0
3 years ago
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Zolol [24]

Answer:

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Explanation:

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3 years ago
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Answer:

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Explanation:

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