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Vinil7 [7]
3 years ago
15

A sample of a compound contains 60.0 g C and 5.05 g H. It’s molar mass is 78.12 g/mol. What is the compound’s molecular formula?

Chemistry
1 answer:
Natalka [10]3 years ago
5 0

Answer:

C6H6

Explanation:

C :     60.0 g*1 mol/12 .01g = 5.00 mol

H:      5.05 g * 1 mol/1.01 g= 5.10 mol

C:H = 5.00 mol : 5.10 mol = 1:1

CH is an empirical formula

Molar mass (12.01 +1.01)=13.02 g/mol

78.12g/mol /13.02 g/mol =6

Molecular formula is the empirical that was taken 6 times (CH)6 = C6H6

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Answer:

the nucleus is made up of protons and neutrons

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2 years ago
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Bismuth has a density of 9.80 g/cm^ 3 . What is the of 3.74 cm ^ 3 of Bi?
Ierofanga [76]

Explanation:

density = mass/ volume

mass= density × volume

mass= 9.80× 3.74

mass= 36.652 g

3 0
3 years ago
Why is it more difficult to remove an
Tom [10]
In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outer most orbital. For example, it would be far easier to take electrons away from the larger element of Ca (Calcium) than it would be from one where the electrons are held tighter to the nucleus, like Cl (Chlorine). Hope this helped a little not the exact answer though :)
3 0
3 years ago
Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls t
Kobotan [32]

Now we know that

Q = mc∆T

Where Q is y energy measured in Joules.

m is the mass measured in grams

c is the specific heat of the substance measured in joule per gram degree Celsius.

∆T is the change in temperature measured in degree Celsius.



Let Q1 be the specific heat of the lighter ball.

c1 be the specific heat of the lighter ball.

m1 be the mass of the lighter ball.

∆T1 be the change in the of the lighter ball.


Let Q2 be the specific heat of the heavier ball.

c2 be the specific heat of the heavier ball.

m2 be the mass of the heavierr ball.

∆T2 be the change in the of the heavier ball.


It has been given that the heat lost, that is Q is the same for both the balls of different mass.Which implies Q1= Q2

Specific heat(c) is the same for both the balls since both are made up of iron. c1=c2


Now heat lost by the lighter ball = heat lost by the heavier ball.

Q1= Q2

m1c1∆T1= m2c2∆T2

Since c1=c2

We get

m1/m2= ∆T2/∆T1

Thus we can say since m2>m1,∆T1> ∆T2.

Now initial temperature of both the balls are 100 degree Celsius.

∆T1 = Final temperature(T1 )-100.

∆T2= Final temperature ( T2)-100

Now since the ∆T1> ∆T2 as arrived from the above equation we can conclude that the final temperature of the ball 1 is greater than that of the ball 2. Since the ball 1 as per our assumption is the lighter ball,the final temperature of the ball which has lighter mass is greater than that of the one having a greater mass.

5 0
3 years ago
Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?
lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

#SPJ4

7 0
1 year ago
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