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3 years ago

Which phrase best describes the salinity of ocean water?

Chemistry

2 answers:

mojhsa [17]3 years ago

8 0

Answer:

Number of grams of mineral salt in one liter of water

Explanation:

Salinity :

It is the concentration of dissolved salts .

Salinity means the saltiness or the dissolved salt component in the water body . It is defined as mass (in gram) of the dissolved matter in the water bodies .

salts found in ocean water are :

Salts of chlorides, sulfates , sodium , magnesium, potassium and calcium

For example : world's oceans salinity is 3.5% parts per thousand (g/L) . It means 3.5 g of salt is present in 1 L of seawater .

Note : Sodium chloride is major salt found in ocean . But it is not the only salt present (other salts are also found) . So D can't be the answer

svp [43]3 years ago

7 0

Answer:

the answer is number of grams of salt in one kilogram of water

Explanation:

i just finished the quiz and i used the other guy's answer and got it wrong and i know you already took this test but this is for the next person who comes across this

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

The pressure of a sample of helium in a 200. ml container is 2.0 atm. If the helium is compressed to a pressure of 40. atm witho

Crank

Answer:

$V_2=10mL$

Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

$P_2V_2=P_1V_1$

In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

$V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{2.0atm*200.mL}{40.atm}\\\\V_2=10mL$

Regards!

5 0

3 years ago

True or Fase. Every substance has to be classified as either acid or base?

Sever21 [200]

Answer:

False

Explanation:

Some substances don't have to be

8 0

3 years ago

g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular

Whitepunk [10]

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

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The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

Where the products are H2O and Ba(NO3)2

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3 years ago

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Keith_Richards [23]

Answer:

yes

Explanation:

7 0

3 years ago