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3 years ago

15

An elevator does 150000 j of work while exerting a force of 1000 n to take people to the top floor of a building. What distance

does the elevator travel?

1 answer:

VLD [36.1K]3 years ago

8 0

Answer:

The distance traveled by the elevator is 150 meters.

Explanation:

Work done is defined as the product of force and distance by whcih an object is displaced by applying that force.

$Work(w)=Force(F)\times Displacement(d)$

We have L

Force applied to lift an elevator = F = 1000 N

Displacement of the elevator or distance traveled by an elevator = d

Amount of work done = W = 150,000 J

$150,000 J=1000 N\times d$

$d=\frac{150,000 J}{1000 N}=150 m$

The distance traveled by the elevator is 150 meters.

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Explanation:

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3 years ago

A student measures out exactly 0.105 g of salicylic acid and runs the experiment as dictated in the lab manual. They obtain 0.11

scoray [572]

<u>Answer:</u> The percent yield of the reaction is 8.10 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylic acid = 0.105g

Molar mass of salicylic acid = 138.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylic acid}=\frac{0.105g}{138.12g/mol}=0.0079mol$

The chemical equation for the formation of aspirin from salicylic acid follows:

$\text{Salicylic acid + Acetic anhydride}\rightarrow \text{Aspirin + Acetic acid}$

By Stoichiometry of the reaction:

1 mole of salicylic acid produces 1 mole of aspirin

So, 0.0076 moles of salicylic acid will produce = $\frac{1}{1}\times 0.0076=0.0076mol$ of aspirin

Now, calculating the mass of aspirin from equation 1, we get:

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.0076 moles

Putting values in equation 1, we get:

$0.0076mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=(0.0076mol\times 180.16g/mol)=1.37g$

To calculate the percentage yield of aspirin, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of aspirin = 0.111 g

Theoretical yield of aspirin = 1.37 g

Putting values in above equation, we get:

$\%\text{ yield of aspirin}=\frac{0.111g}{1.37g}\times 100\\\\\% \text{yield of aspirin}=8.10\%$

Hence, the percent yield of the reaction is 8.10 %.

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3 years ago

Rank cesium, chlorine, and chromium in order of increasing electron affinity

Fantom [35]

Answer:

Cs, Cr, Cl

Explanation:

The Placement Of The Elements Increase In Electro Affinity.

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3 years ago

Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the g

nasty-shy [4]

Answer:

5' RG GWCCY 3'

3' YCCWG GR 5'

Explanation:

The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).

The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'

target Sequence: 5' RGGWCCY 3'

3' YCCWGGR 5'

The enzyme cleavage point is at:

5' RG^GWCCY 3'

3' YCCWG^GR 5'

The product of the cleavage will give a sticky end Cleavage:

5' RG GWCCY 3'

3' YCCWG GR 5'

Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.

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3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

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Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

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3 years ago