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3 years ago

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Chemistry

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

P * V = k

To obtain the proportionality factor k you must make the quotient:

$k=\frac{V2}{V1} =\frac{6 L}{2 L}$

k= 3

This means that if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

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The size of the gas particles compared to the overall volume of the gas is 22.4 L

jeka57 [31]

Well. Yeah. Basically.

1 mole of any gas is 22.4 L.

Not really sure if that answers your question but that's what i know.
Anyway if we are talking about the size then usually gases are filled with empty space.

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4 years ago

What mass of magnisium bromide is formed when 1.00 g of Mg + br

ella [17]

Mass of Magnesium bromide : 5.76 g

<h3>Further explanation</h3>

Complete question

What mass of Magnesium bromide is formed when 1.00 g of magnesium reacts with 5.00 g of bromine?

Reaction

$\tt Mg+Br_2\rightarrow MgBr_2$

mol Mg

$\tt \dfrac{1}{24}=0.042$

mol Br₂

$\tt \dfrac{5}{160}=0.0313$

Limiting reactants : Br₂(smaller)

mol MgBr₂ = mol Br₂=0.0313

mass MgBr₂ :

$\tt 0.0313\times 184,113 =5.76~g$

5 0

4 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

True or False questions

Vesnalui [34]

Answer:

I did not understand the question

3 0

3 years ago

Be sure to answer all parts.

MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol$

Now, molarity of ethanol solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M$

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol$

Now, molarity of sucrose solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M$

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol$

Now, molarity of sodium chloride solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M$

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

4 0

3 years ago