All categories

2 years ago

Why do oceanographers take measurements of the conductivity of the

Chemistry

1 answer:

bazaltina [42]2 years ago

8 0

Answer:

B. Conductivity can be used to calculate the salinity of the water.

I am 100% sure this is the correct answer :)

You might be interested in

Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlig

fomenos

Answer:

The one left in the hot sunlight.

Explanation:

The solubility of gases decreases when temperature increases. The gas in the soda pop (CO2) left in the sun will not stay dissolved as much as the on left in the refrigerator.

3 0

2 years ago

HELPP!!!!!!!!!!

muminat

1. Mg + CuSO4 —> MgSO4 + Cu

=> Single displacement reaction

2. 2HCI + Mg(OH)2 —> MgCI2 + H2O

=> Acid - base reaction

3.AgNO3 + NaCI —> AgCI + NaNO3

=> Double displacement reaction

8 0

3 years ago

This image shows a...

likoan [24]

Answer:

I think is the last one!!!! punnet square ⬜

5 0

2 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

5 0

3 years ago

Identify each of the following as either an element or Compound. Write (E) for Element and (C)

castortr0y [4]

Carbon = C

Water = C

Aluminum foil = E

Plastic = E

Tin = E

Silicon dioxide = C

Helium = C

Arsenic = C

Carbon dioxide = C

Sodium Chloride = C

4 0

3 years ago