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Ede4ka [16]
3 years ago
10

Give examples of simple machines advantage and disadvantage

Chemistry
1 answer:
ra1l [238]3 years ago
6 0
Simple machines help reduce the amount of work being done, as they help by using a simple function. One disadvantage is that they do not help completely. 
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Monobromination of toluene gives a mixture of three bromotoluene products. Draw and name them.
Art [367]

Answer:

<em>o</em>-bromotoluene, <em>m</em>-bromotoluene  and <em>p</em>-bromotoluene.

Explanation:

Hello,

In this case, on the attached picture you will find the reaction which yields <em>o</em>-bromotoluene as the first product, <em>m</em>-bromotoluene as the second product and <em>p</em>-bromotoluene as the last one since the substitution could be done at the second (ortho), third (meta) or fourth (para) carbons on the toluene.

Regards.

5 0
3 years ago
You have 0.21 moles of Al how many atoms do you have
ZanzabumX [31]

Number of atoms : 1.26 x 10²³

<h3>Further explanation  </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

0.21 moles of Al, so n = 0.21

Number of atoms :

\tt N=0.21\times 6.02\times 10^{23}\\\\N=1.26\times 10^{23}

8 0
2 years ago
Which one of the following represents the number of units of each substance
netineya [11]

Answer:B

Explanation:Coefficient represents the number of units of each substance.

4 0
3 years ago
For the reaction Fe3O4(s) + 4H2(g) --&gt; 3Fe(s) + 4H2O(g)
mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = 151.2 kJ = 151200 J

\Delta S^o = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(151200J)-(328.0K\times 169.4J/K)

\Delta G^o=95636.8J=95.6kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = 95636.8 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k

k=1.70\times 10^{15}

Therefore, the value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

3 0
3 years ago
I need help with this question plz ASAP
Orlov [11]

Answer:

D

Explanation:

i farted

7 0
2 years ago
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