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Archy [21]
3 years ago
11

How much work is required to lift a 0.500 kg block 18.5 m?

Chemistry
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

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What reaction conditions most effectively conver a cabocxylic acid to a methly ester?
VLD [36.1K]

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the <u>esterification reaction</u>. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).  

In this case, we need the <u>methyl ester</u>, therefore we have to choose the <u>appropriate alcohol</u>, so we have to use the <u>methanol</u> as reactive if we have to produce the methyl ester.

7 0
3 years ago
An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its
german
Hello!

First you need to calculate q 
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>

<span>q = m x c x delta T </span>

<span>7211 J = 80.0 g x c x (225-25) °C </span>

<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
5 0
3 years ago
What is advantage of using single fixed pulley although it doesnot multiply force?​
luda_lava [24]

A single fixed pulley is used to lift load by changing the direction of the lifting force. Even though a single fixed pulley has no mechanical gain in that, the effort applied is greater than the load, it is preferred in use since you do not have to pull or push the pulley up and down.

4 0
2 years ago
Read 2 more answers
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
Pls do it I don’t understand
Ann [662]
The first question's answer is

6 0
3 years ago
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