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3 years ago

Fill in the blank spot.

Chemistry

1 answer:

Anastasy [175]3 years ago

6 0

Answer:

Alloys and Composites

Explanation:

Drill bits are made of alloys because alloys are very much resistant to corrosion and are created when one or multiple metals are mixed with other elements that are non-metallic in nature. Due to their greater strength, they are used in the power drills in order to create borehole drills at the mining site. Borehole drilling is commonly done in order to understand the sub-surface strata before mining.

The bodies along with the handles of power drills are constructed of composite insulators because within the composites there is the absence of free electrons that has the ability to flow and produce electricity. So they are considered to be used as a good insulator.

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Answer:

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A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume

Soloha48 [4]

Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then

PV = c

Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,

P₁V₁ = P₂V₂

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1.00 atm x 3.60 L = 2.50 atm x y

We cleared y to find V₂,

V₂ = y = $\frac{1.00 atm x 3.60 L}{2.50 atm}$ = 1.44 L

Then, the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

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3 years ago

Limitation for osmosis experiment

taurus [48]

What was the question here?

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2 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

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