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OlgaM077 [116]
3 years ago
9

to find the density of stopper I weighted it and found its mass to 4.8g. After that I filled a graduated cylinder with 32.1mL of

water. After adding the stopper, the water level rose to 39.2mL. What is the density of the stopper?
Chemistry
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer:

0.68g/ml

Explanation:

The density of an object is its mass per unit volume. It is calculated using the formula

Density = mass / volume

Mass of stopper weighed = 4.8g

The volume of stopper can be got by subtracting the (volume of water) from the (volume of water+stopper) i.e.

= 39.2ml - 32.1ml

= 7.1ml

Volume of stopper = 7.1ml

Density of stopper= 4.8/7.1

Density= 0.676056

Therefore, the density of the stopper is 0.68g/ml

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What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?
insens350 [35]
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
6 0
3 years ago
What is the net ion charge of Beryllium?​
KATRIN_1 [288]

Explanation:

Beryllium is a group 2 element and its atomic number is 4. Electronic configuration of beryllium is 1s^{2}2s^{2}.

Since, a beryllium contains two valence electrons so, in order to attain stability it will readily lose its 2 valence electrons.

Therefore, a beryllium atom upon losing two valence electrons will acquire a +2 charge.

Thus, we can conclude that the net ion charge of Beryllium is +2.

7 0
3 years ago
A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i
stepladder [879]

Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

<em>Where P is pressure: 4.0atm</em>

<em>V is volume: 9.6L</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 300K</em>

<em>And n are moles of the gas</em>

<em />

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is:

<h3>C) 1.3 mol</h3>

6 0
2 years ago
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Of the elements fe li te u and he which are considered group b elements
VLD [36.1K]
A group B element is Fe 
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3 years ago
How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

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

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0
3 years ago
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