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3 years ago

Each particle of water in an ocean wave moves in a circle true or false

Chemistry

1 answer:

djverab [1.8K]3 years ago

8 0

Answer:

false

Explanation:

when the particles get in the water they just simply vibrate

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What is the temperature of 0.750 mol of a gas stored in a 6,050 mL cylinder al 221 atm?

Rasek [7]

Answer:

T = 246 K

Explanation:

Given that,

Number of moles, n = 0.750 mol

The volume of the cylinder, V = 6850 mL = 6.85 L

Pressure of the gas, P = 2.21 atm

We need to find the temperature of the gas stored in the cylinder. We know that,

PV= nRT

Where

R is gas constant

T is temperature

So,

$T=\dfrac{PV}{nR}\\\\T=\dfrac{2.21\times 6.85}{0.75\times 0.0821}\\T=245.85\ K$

T = 246 K

So, the temperature of the gas is equal to 246 K.

4 0

2 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

3 years ago

An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga

Scorpion4ik [409]

Answer:

51207 torr is the new pressure of the gas

Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>

Computing the values of the problem:

P1 = 710torr

V1 = 5.0x10²mL

T1 = 273.15 + 30°C = 303.15K

P2 = ?

V2 = 25mL

T2 = 273.15 + 820°C = 1093.15K

Replacing:

710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K

P2 = 51207 torr is the new pressure of the gas

4 0

2 years ago

Write the two balanced equations, one for the incomplete combustion of methane (to form carbon monoxide and water vapor), and an

kompoz [17]

Answer:

$2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\$

Explanation:

$2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\$

The incomplete combustion of alkanes, and other fuels actually, happens when there is a limited supply of oxygen. Instead of the fuel burning completely to produce carbon dioxide, it produces carbon monoxide instead.

This gas is harmful to jumans because it combines with haemoglobin in lood and takes up space that belongs to oxygen which can lead to suffocation or even death

6 0

2 years ago

Which state of matter has a constant volume, but changes shape?\?

never [62]

Liquids stays the same volume but the bonds are spaced out enough that it can take the shape of whatever container it’s in.

8 0

3 years ago