What metal is easily oxidized in rocks?

The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol

Thepotemich [5.8K]

The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O

M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5

moles of HCl = 5.5/37.5=0.14

M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)

M(C2H6O)= 2x12+ 6 + 16=46,

moles of C2H6O=200g / 46 =4.35 moles

the sum of the moles is 0.14 + 4.35 = 4.501 moles

finally, The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031

because it can be found by 0.14 / 4.501= 0.031

5 0

3 years ago

How many electrons are contained in an Au3+ ion?(1) 76 (3) 82 (2) 79 (4) 197

Maurinko [17]

It is 79 - + 3 = 76 electrons.

5 0

3 years ago

HELP ASAP ILL GIVE BRAINLIEST!!

Dahasolnce [82]

Of course we would experience them.

3 0

2 years ago

If the distance between two objects is decreased to - of the original

Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

$F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R$

And the force at this distance would be written in terms of the same equation:

$F_2 = G\frac{m_1m_2}{R_2^2}$

Find the ratio between the final and the initial force:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}$

Substitute the value for the final distance in terms of the initial distance:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}$

Simplify:

$\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}$

This means the new force will be \frac{1}{100} of the original force.

8 0

3 years ago

The entropy change for a real, irreversible process is equal to:______.

taurus [48]

Answer:

The entropy change for a real, irreversible process is equal to zero.

The correct option is 'c'.

Explanation:

Lets look around all the given options -:

(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

Hence , the correct option is 'c' that is zero.

8 0

3 years ago