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3 years ago

How many moles of methanol must be burned to release 1454 KJ of energy

Chemistry

1 answer:

ra1l [238]3 years ago

6 0

A lot of moles ya know but moles need sticks. Big sticks equals big moles and big moles equal for me 5 points

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Identify which of the following two reactions you would expect to occur more rapidly: (1) addition of HBr to 2-methyl-2-pentene

arlik [135]

Answer:

(1) addition of HBr to 2-methyl-2-pentene

Explanation:

In this case, we will have the formation of a carbocation for each molecule. For molecule 1 we will have a tertiary carbocation and for molecule 2 we will have a secondary carbocation.

Therefore the most stable carbocation is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)

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3 years ago

One mole of copper(ii) sulfate, cuso4, contains ________ o atoms.

Ierofanga [76]

According to avogadro constant, the number of units in one mole of any substance contain 6.022 x10 ^23 atoms

therefore the number of o atoms in one mole of CuSO4 = 6.022 x 10 ^ 23

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3 years ago

Read 2 more answers

Andrea and the other members of the tennis team are practicing hitting tennis balls. If two people hit identical balls at the sa

almond37 [142]

Answer:

Explanation:

The formula for calculating kinetic energy is 1/2mv^2, where m is the mass and v is the velocity or speed of the object. Since all of the balls have the same mass, the one with the highest speed or velocity will have the highest kinetic energy. Hope this helps!

8 0

3 years ago

An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr

Lera25 [3.4K]

Answer:

$\sigma_y\ =210.2\ MPa$

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

$\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}$

$\sigma_y\ =Yield\ strength$

d = diameter

K =Constant

$\sigma_o\ =material\ constant$

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

$163=\sigma_o+\dfrac{K}{\sqrt 35}$ ------------1

d= 17 μm ,yield strength = 192 MPa

$192=\sigma_o+\dfrac{K}{\sqrt 17}$ ------------2

From equation 1 and 2

$192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}$

K=394.53

By putting the values of K in equation 1

$163=\sigma_o+\dfrac{394.53}{\sqrt 35}$

$\sigma_o\ =96.31\ MPa$

$\sigma_y=96.31+\dfrac{394.53}{\sqrt d}$

Now when d= 12 μm

$\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}$

$\sigma_y\ =210.2\ MPa$

4 0

3 years ago

If you have 222 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?

Dima020 [189]

If the half life is 5 days, you would have to divide 222 by 2 3 times.
so 222/2 =111g
111/2 = 55.5g
55.5/2 = 27.75g

after four half life the mass would be
27.75/2
13.875g

the percent would be
13.875/222
= 6.25%

7 0

3 years ago

How many moles of methanol must be burned to release 1454 KJ of energy￼

How many moles of methanol must be burned to release 1454 KJ of energy