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3 years ago

The measure of the amount of dissolved salt in a liquid sample is called

2 answers:

6 0

The amount of dissolved salt in the liquid sample is measured and reported as salinity. The salinity is usually measured in parts per thousand (ppt). The salinity of ocean averages 35 ppt while that of the river averages 0.5 ppt or less. In other terms, the word salinity is the saltiness.

vladimir2022 [97]3 years ago

3 0

The correct answer is C. The measure of the amount of dissolved salt in a liquid sample is called sallinity. It defiens how much salty is a solution or how much amount of salt is present in the solution. It is measured in thousand part per gram of the sea water.

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What is a unifying principle that explains a body of

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3 years ago

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

$W = -P*(V_{2} - V_{1})$

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

$V_{1}$ is the initial volume = 0.350 L

$V_{2}$ is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( $V_{2}$ - 0.350) L

$V_{2}$ = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

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3 years ago

Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo

Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess. There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

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3 years ago

What is the total number of atoms in this formula? NaHCO3

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6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.

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