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NNADVOKAT [17]
3 years ago
5

A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos

ition of fluorine
Chemistry
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

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Which gas variable has to do with the amount of collisions of gas particles has?
Firlakuza [10]
The answer is temperature
3 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
what is the balanced equation for the reaction of solid magnesium metal with oxygen gas to produce solid magnesium oxide
GaryK [48]

Answer:

2Mg + O₂ ⟶ 2MgO

Explanation:

Step 1. Start with the most complicated-looking formula (O₂?).

Put a 1 in front of it.

Mg + 1O₂ ⟶ MgO

Step 2. Balance O.

We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.

Mg + 1O₂ ⟶ 2MgO

Step 3. Balance Mg.

We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.

2Mg + 1O₂ ⟶ 2MgO

Every formula now has a coefficient. The equation should be balanced. Let’s check.

<u>Atom</u>  <u>On the left</u>  <u>On the righ</u>t

  Mg             2                  2

  O                2                  2

All atoms are balanced.

The balanced equation is

2Mg + O₂ ⟶ 2MgO

4 0
3 years ago
How many significant figures are in 0.0233?
viva [34]

Answer:

there are 3 significant figures

Explanation:

do not count the first 2 zeros. only the nimbers after the zero

7 0
4 years ago
2. Incoming wastewater, with BOD5 equal to 200 mg/L, is treated in a well-run secondary treatment plant that removes 90 percent
Lesechka [4]

Answer:

10.8 ml

Explanation:

The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.

See attached file

5 0
3 years ago
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