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vodomira [7]
3 years ago
8

1. Kaleigh notices when she goes to the beach that sometimes the water rises as high as the pier. At other times of the day, the

water barely covers the pillars under the pier. These differences in water level are primarily due to the gravitational influence of which of the following?
a. The Earth’s revolution b. The Moon c. asteroids d. comets
Chemistry
1 answer:
Nataly [62]3 years ago
5 0
B. The moon
Tides are due to the gravitational forces of the Moon and the Sun. The tide is high when the moon or the sun or both are on the side of the sea and vice versa.
The effect of tides due to the moon is much greater than that of the sun, because the moon is much closer to the earth.
You might be interested in
Draw the alcohol that is the product of the reduction of 3-methylpentanal.
natali 33 [55]

Answer: The product from the reduction reaction is

CH3-CH2-CH(CH3)-CH2-CH2OH

IUPAC name; 3- Methylpentan-1-ol

Explanation:

Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.

Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal

CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;

3-methylpenta-1-ol .

The structure of the product is:

CH3-CH2-CH(CH3)-CH2-CH2OH

7 0
3 years ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
nikdorinn [45]

Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

\text{Moles of }Al=\frac{55g}{27g/mol}=2mole

Now we have to calculate the moles of O_2.

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of O_2

So, 2 mole of Al react with \frac{3}{4}\times 2=0.75\times 2 moles of O_2

Now we have to calculate the volume of O_2 consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of O_2 occupies 22.4 liter volume of O_2 gas

So, 0.75\times 2 mole of O_2 occupies 0.75\times 2\times 22.4 liter volume of O_2 gas

Therefore, the volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

3 0
3 years ago
Recently, the United States government decided to construct a huge crypt in the middle of Yucca Mountain in Nevada to bury high-
Anton [14]
The three concerns that the residents of this area might be:

1) <span>The cost of not moving forward is extremely high, so they opposed the plan, as they think it would affect US economy

2) </span><span>Nuclear waste disposal capability is an environmental imperative, so their environment would be polluted by very radioactive materials.

3) </span>Demand for new nuclear plants also demands disposal capability which supports national security but again, their site will be no longer for them. But unfortunately, <span>Extensive studies consistently show Yucca Mountain to be a sound site for nuclear waste disposal so the plan can't be abolished.

Hope this helps!</span>
4 0
4 years ago
Read 2 more answers
Can someone help me plz
LenKa [72]

Answer:

See Explanation

Explanation:

The answer should be 8 mL according to my calculations!

7 0
4 years ago
) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react
Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of B_5H_9 with oxygen gas follows:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

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

\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ

To calculate the amount of heat released for the given amount of B_5H_9(l), we use unitary method, we get:

When 2 moles of B_5H_9(l) reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of B_5H_9(l) will react, the amount of heat released will be = \frac{5259}{2}\times 0.211=554.8kJ

Hence, the amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

7 0
3 years ago
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