True or false

When the pressure on 2.5 liters of oxygen is decreased from 2.0 atm to 1.0 atm, the volume of the gas

uranmaximum [27]

Answer:

volume of the gas is 5.0L

Explanation:

Using Boyle's law that state the pressure of a gas is inversely proportional to volume of it occupies when temperature is constant, it is possible to write:

P₁V₁ = P₂V₂

Where P is pressure, V is volume and 1 and 2 are initial and final states.

If initial volume is 2.5L, initial pressure is 2.0atm and 1.0atm is final pressure, final volume is:

2.0atm*2.5L = 1atm V₂

5.0L = V₂

Thus, volume of the gas is 5.0L.

5 0

3 years ago

When convex mirrors are used for security purposes, the images they form

konstantin123 [22]

Answer:false

Explanation:

3 0

3 years ago

Suppose you are a food chemist working for a company that makes and manufactures soda. Your job is to create a new soft drink wi

Mekhanik [1.2K]

Answer:

The answer to your question is given after the questions so I just explain how to get it.

Explanation:

a)

Get the molecular weight of Phosphoric acid

H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)

= 3 + 31 + 64

= 98 g

98 g ----------------- 1 mol

0.045 g --------------- x

x = (0.045 x 1) / 98

x = 0.045 / 98

x = 0.00046 moles or 4.6 x 10 ⁻⁴

b)

Molarity = $\frac{moles}{volume}$

Molarity = $\frac{0.00046}{0.35}$

Molarity = 0.0013 or 1.31 x 10⁻³

c)

Formula C₁V₁ = C₂V₂

V₁ = C₂V₂ / C₁

Substitution

V₁ = (0.0013)(1) / 0.01

Simplification and result

V₁ = 0.0013 / 0.1

V₁ = 0.13 l = 130 ml

7 0

3 years ago

The three isomers of pentane have different

8_murik_8 [283]

The three isomers of pentane have different structural formulas.

8 0

3 years ago

Read 2 more answers

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$