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2 years ago

What is the mass in grams of 85.32 mL of blood plasma with a density of 1.03 g/mL

1 answer:

3 0

Well if you're looking for grams, all you need to do is cancel out units. (ml)(g/ml)=g because the ml cancels out. Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :) Good luck! Hope that helped. When in doubt, look at the units.

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(HELP ME ASAP!)

slega [8]

A) Ca(OH)2 + CO2 —> CaCO3 + H2O

B) when Ca(OH)2 is reacted with CO2, the CaCO3 produced is a precipitate which turns the solution milky

8 0

2 years ago

What are the base units the SI units are based on? A-10 B-100 C-1,000 D-1

Olegator [25]

Answer:

A-10

Explanation:

In the SI, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth.

IM NOT SURE PO

4 0

2 years ago

A sample of gas contains 0.1700 mol of NH3(g) and 0.2125 mol of O2(g) and occupies a volume of 17.8 L. The following reaction ta

telo118 [61]

Answer:

The volume of the sample after the reaction takes place is 19.78 L.

Explanation:

The given variables are;

Number of moles of NH₃(g) = 0.1700 mol

Number of moles of O₂(g) = 0.2125 mol

Volume occupied by the mixture = 17.8 L

The reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Then takes place

That is 4 moles of NH₃(g) reacts with 5 moles of O₂(g) to produce 4 moles of NO(g) and 6 moles of H₂O(g).

Since there are less number of moles of NH₃(g) (= 0.1700 mol) in the mixture, we factor the above equation by the number of moles of NH₃(g) present.

That is,

1 moles of NH₃(g) reacts with 5/4 moles of O₂(g) to produce 1 moles of NO(g) and 3/2 moles of H₂O(g).

Therefore,

0.1700 mol of NH₃(g) reacts with 5/4×0.1700 moles of O₂(g) to produce 0.1700 moles of NO(g) and 3/2×0.1700 moles of H₂O(g).

Which gives

0.1700 mol of NH₃(g) reacts with 0.2125 moles of O₂(g) to produce 0.1700 moles of NO(g) and 0.255 moles of H₂O(g).

Therefore, all of the NH₃(g) and O₂(g) are consumed in the reaction and the present gases in sample then becomes

0.1700 moles of NO(g) and 0.255 moles of H₂O(g).

Total number of moles of reactant = 0.17 + 0.2125 = 0.3825

Total number of moles of product formed = 0.17 + 0.255 = 0.425

However, Avogadro's law states that equal volume of all gases at the same temperature and pressure contains equal number of molecules.

That is volume occupied by 0.3825 moles of gas = 17.8 L

Therefore the volume occupied by 0.425 moles of gas = 17.8×0.425/0.3825 L = 19.78 L

3 0

3 years ago

Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0

Alexxx [7]

Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol

We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

 H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed

4 0

2 years ago

Describe what you would expect to observe, if anything, when a piece of nickel is placed in a solution of copper(II) nitrate. (s

Tom [10]

When a piece of nickel is placed in a solution of copper(II) nitrate:

a. Solid copper would form on the surface of the nickel metal.

c. Nickel atoms are converted to nickel ions in solution.

d. The blue color of the solution would become lighter.

Let's consider what happens when a piece of nickel is placed in a solution of copper(II) nitrate. Cu has a higher reduction potential than Ni, thus Cu will reduce and Ni will oxidize. This will be a single displacement redox reaction. The corresponding balanced molecular equation is:

Ni(s) + Cu(NO₃)₂(aq) ⇒ Ni(NO₃)₂(aq) + Cu(s)

The net ionic equation is:

Ni(s) + Cu²⁺(aq) ⇒ Ni²⁺(aq) + Cu(s)

Describe what you would expect to observe, if anything, when a piece of nickel is placed in a solution of copper(II) nitrate.

a. Solid copper would form on the surface of the nickel metal. YES. Solid copper will deposit on the surface of nickel metal.

b. The blue color of the solution would become darker. NO. The blue color, caused by Cu²⁺ ions, will become lighter, as their concentration decreases.

c. Nickel atoms are converted to nickel ions in solution. YES. Ni is oxidized to Ni²⁺.

d. The blue color of the solution would become lighter. YES. Due to the decrease in the concentration of Cu²⁺ ions.

e. No reaction would occur. NO.

When a piece of nickel is placed in a solution of copper(II) nitrate:

a. Solid copper would form on the surface of the nickel metal.

c. Nickel atoms are converted to nickel ions in solution.

d. The blue color of the solution would become lighter.

You can learn more about redox reactions here: brainly.com/question/13978139

6 0

1 year ago