The galaxy we live in is the Milky Way galaxy, and it is a spiral galaxy. Galaxies contains billions of which of the following?

What do lightning and stars have in common?

irina1246 [14]

2... one though it is true is not very strong

three is very incorrect and dumb

four... idek

4 0

2 years ago

How many molecules of sulfur dioxide are present in 1.60 moles of sulfur dioxide

ELEN [110]

For this question, you must know that there are 6.022e23 atoms/molecules per mole of any substance (this is Avogadro's number). Therefore, your answer is 6.022e23 * 1.60 = 9.64e23 molecules of sulfur dioxide. (the "e" represents "times ten to the power of ___ ")

5 0

3 years ago

What is the difference in concentration between a pH of 7 and 12?

Ratling [72]

Answer:

The pH of a solution is simply a measure of the concentration of hydrogen ions,

H

+

, which you'll often see referred to as hydronium cations,

H

3

O

+

.

More specifically, the pH of the solution is calculated using the negative log base

10

of the concentration of the hydronium cations.

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

pH

=

−

log

(

[

H

3

O

+

]

)

a

∣

−−−−−−−−−−−−−−−−−−−−−−−−

Now, we use the negative log base

10

because the concentration of hydronium cations is usually significantly smaller than

1

.

As you know, every increase in the value of a log function corresponds to one order of magnitude.

Explanation:

4 0

2 years ago

Read 2 more answers

You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water = 600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = $\frac{C_1}{C_2}$

where; $C_1=$ solubility of compound in the organic solvent and $C_2$ = solubility in aqueous water.

So; we can represent our data as:

$K=(\frac{A_{(g)}}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL} )$

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6= $(\frac{0.6-B(mg)}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL})$

4.6 = $\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}$

4.6 × $(\frac{B_{(mg)}}{100mL})$ = $(\frac{0.6-B(mg)}{60mL} )$

4.6 B $*\frac{60}{100}$ = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = $\frac{0.6}{3.76}$

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0

2 years ago

Read 2 more answers

(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n

quester [9]

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

$5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O$

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene = $\frac{0.03 g}{78 g/mol}=0.0003846 mol$

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

$\frac{15}{5}\times 0.0003846 mol=0.0011538 mol$ of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

4 0

2 years ago