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Kamila [148]
2 years ago
11

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.
Chemistry
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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We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
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Solve for P2 on your calculator and that should get you to the answer.


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Draw the product obtained when trans-2-butene is treated first with br2 in ch2cl2, second with nanh2 in nh3, and then finally wi
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<span>In order to do this, you have change the alkene into an alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions. -NH2 will eliminate an H from carbons 3 and 4. This double elimination will make the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The final product will be CIS-3-octene.</span>

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What atomic or hybrid orbitals make up the bond between c2 and o in acetaldehyde, ch3cho?
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Thus for a c = O bond, the atomic orbital configuration is sp3 containing 1 s orbital and 2 pi bonds. 

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2 years ago
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

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Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
2 years ago
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