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2 years ago

Why does spring tides occur

Chemistry

1 answer:

g100num [7]2 years ago

8 0

Answer:

Because the Moon and the Sun are along the same line with the Earth

Explanation:

Tides are phenomenon that occur periodically on Earth. The term "tides" refer to the increase/decrease in the sea level in oceans and seas due to the combined effect of the gravitational force of the Moon and of the Sun on the water on Earth. In fact, the Moon (and the Sun also) exerts an attractive gravitational force on the water in the oceans, causing the sea level to rise when the Moon is facing that portion of water.

In particular, two types of tides are possible:

- Spring tides: these occurs when the Moon and the Sun are along the same line. In this situation, the gravitational forces exerted by the Moon and the Sun on the water act along the same line, so the combined effect is to increase the amount of tide, so the sea level rises more.

- Neap tides: these occurs when the Moon and the Sun are at 90 degrees from each other, with respect to the Earth. As a result, the gravitational force exerted by the Moon and the Sun on the water act perpendicular to each other, so the combined effect of the tide on the water is less than the spring tides, so the sea level is lower than the spring tides.

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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

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

Which of these form an ionic bond

zhenek [66]

Answer:

the second 1

Explanation:

7 0

2 years ago

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

tensa zangetsu [6.8K]

<h3>Answers:</h3>

1) 2 Units of Ozone

2) 3 Units of Ozone

3) 9 Units of Ozone

<h3>Solution:</h3>

1) From 6 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

6 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

9 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 3 Units of Ozone

3) From 27 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

27 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 9 Units of Ozone

3 0

3 years ago

Which of the following leads to a higher rate of diffusion?

Bingel [31]

I thought the answer is d

5 0

2 years ago

Read 2 more answers