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hichkok12 [17]
3 years ago
9

How many moles of solute are in 5 liters of a 5 M solution?

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
6 0
500 because you need the
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The picture below is a great example of __________ biodiversity.
ss7ja [257]
Low because it is not diverse. It is just seed corn.
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3 years ago
Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger
konstantin123 [22]
The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.
6 0
3 years ago
0.00000000082 -<br> scientific notation
Nadusha1986 [10]

Answer:

8.2 x 106^-11

Explanation:

To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2

Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.

Once the decimal is in the right spot count the spots moved.

Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.

6 0
3 years ago
(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would
Lelu [443]

Answer:

(c) The retention time would be higher (d) The retention time would be lower.

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0
3 years ago
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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