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uranmaximum [27]
2 years ago
13

Describe the pattern observed in volume measurements as different masses or volumes of the unknown solid or liquid were selected

. Explain why this pattern exists.
Chemistry
1 answer:
ahrayia [7]2 years ago
7 0

Answer:

Throughout the expansion segment below, the definition including its instance specified is defined.

Explanation:

  • Volume is the sum of space filled by measurement of material throughout 3 directions. The volume of either a compound is mainly determined by the amount and the process including its molecules.
  • The larger the volume as well as strength including its solid as well as liquid, the greater the total weight.

That's because the combined volume or total weight is supposed to have been significantly greater. After all, there is so much mass.

You might be interested in
Henri becquerel and the curies found out that _____.
natulia [17]

Henri becquerel and the curies found out that atoms were not indivisible and indestructible

Explanation:

The French Physicist, Henri Becquerel, created the appearance of radioactivity. He shared the Nobel Prize in Physics with Marie Curie and Pierre Curie (Marie's husband) in 1903 for their work in radiation. A radionuclide will emit transmission through the process of radioactive sense.

5 0
2 years ago
Read 2 more answers
Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix
Genrish500 [490]

Answer :

(a) Limiting reactant = LiH

(b) The excess reactant = BCl_3

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of B_2H_6 or percent conversion of LiH to B_2H_6 is, 38.80 %

(e) The mass of LiCl produced is, 1066.42 lb

Explanation : Given,

Mass of LiH = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of BCl_3 = 1000 lb = 453592 g

Molar mass of LiH = 7.95 g/mole

Molar mass of BCl_3 = 117.17 g/mole

Molar mass of B_2H_6 = 27.66 g/mole

Molar mass of LiCl = 42.39 g/mole

First we have to calculate the moles of LiH and BCl_3.

\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles

\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

6LiH+2BCl_3\rightarrow B_2H_6+6LiCl

From the balanced reaction we conclude that

As, 6 moles of LiH react with 1 mole of BCl_3

So, 11411.13 moles of LiH react with \frac{11411.13}{6}=1901.855 moles of BCl_3

From this we conclude that, BCl_3 is an excess reagent because the given moles are greater than the required moles and LiH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant (BCl_3).

\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100

\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of B_2H_6.

As, 6 moles of LiH react to give 1 mole of B_2H_6

So, 11411.13 moles of LiH react to give \frac{11411.13}{6}=1901.855 moles of B_2H_6

Now we have to calculate the mass of B_2H_6.

\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6

\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g

Now we have to calculate the percent yield of B_2H_6.

\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%

The percent yield of B_2H_6 or percent conversion of LiH to B_2H_6 is, 38.80 %

Now we have to calculate the moles of LiCl.

As, 6 moles of LiH react to give 6 mole of LiCl

So, 11411.13 moles of LiH react to give 11411.13 moles of LiCl

Now we have to calculate the mass of LiCl.

\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl

\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb

The mass of LiCl produced is, 1066.42 lb

8 0
2 years ago
Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben
sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O →  12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0
2 years ago
With the help of balanced chemical equation explain what happens when:(1)Zinc is put in dilute hydrochloric acid (2)Zinc is put
UkoKoshka [18]
Find the attached document.

6 0
2 years ago
Read 2 more answers
How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?
VARVARA [1.3K]
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
4 0
2 years ago
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