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2 years ago

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In a post office, a 3-m long ramp is used to move carts onto a dock that is higher than 1 m. Which describes how the IMA of this

ramp compares with the IMA of the ramp that has a height of 1 m? It is greater. It depends on friction. It is the same. It is less.

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

well I think the answer is it depends on the friction

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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

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Explanation:

South America land mass serves as a deflector for the South equatorial current. This deflection the current causes them to move in a different direction. If the Continent were not present the direction of the South equatorial current would not change and it would continue to flow in the west.

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Which factor explains why coal dust in an enclosed space is more explosive than coal dust blown outdoors into an open space ? A.

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The correct option is B.

Coal dust refers to the powered form of coal. Because of the high surface area of coal dust it is highly prone to dust explosion, which involves rapid combustion of fine particles that are suspended in the air; this usually occur in an enclosed place. Coal dust in an enclosed place is more explosive than coal dust that is blown outdoor in an open space because the coal dust in an enclosed place is more concentrated due to restricted space, thus it is more liable to explosion.

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Answer:

The concentration is 0,2925M

Explanation:

We use the formula

C initial x V initial = C final x V final

11,7 M x 25 ml = C final x 1000 ml

C final= (11,7 M x 25 ml)/1000 ml = 0, 2925 M

(This formula applies to liquid solutions)

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The change is that the air goes up then forms clouds.

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