You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.
<em>Step 1.</em> Calculate the volume of 90 % alcohol needed
You can use the dilution formula
<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2
where
<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions
<em>C</em>1 and <em>C</em>2 are the concentrations
You can solve the above formula to get
<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2
<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %
V2 = ?; _____<em>C</em>2 = 90 %
∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL
You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA
<em>Step 2</em>. Calculate the amount of water to add.
Volume of water = 270 mL – 195 mL = 75 mL
To make 1 Molar solution of hemoglobin ; 1600 grams of hemoglobin will be dissolved in 1 liter of water
The molecular weight of Hemoglobin is approximately 16,000 Daltons, when hemoglobin is converted to mM
16000 Dalton = 16000 ( g/mol )
given that 1 Dalton = 1 g/mol
To make 1 molar solution of hemoglobin using 1 liter of water
1 liter = 1000 grams
16000 Dalton = 16000 g/mol
Hence 16,000 grams of Hemoglobin is required to make 1 Molar solution of hemoglobin using 1 liter of water.
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Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
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