Ask question

Ask question

All categories

3 years ago

14

An unknown element, X, has an atomic mass of 107.868 amu. The X-109 isotope (108.905 amu) is 48.16%. What is the amu of the othe

r isotope (report final answer to the correct number of significant figures)

1 answer:

juin [17]3 years ago

4 0

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

X-109 = 108.905amu, i = 48.16% = 0.4816

X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

106.905 = X-107

<h3>106.905 amu is the mass of the other isotope</h3>

You might be interested in

11. Which of the following accurately describes properties of valence? A. Nonmetallic elements tend to have a positive valence a

marin [14]

Atoms, the main constituents of matter, consist of positively charged protons and neutral neutrons within a nucleus which are surrounded by a sea of electrons that sit in distinct shells. The electrons on the outer shell are known as valence electrons. The valence can be descibed as the smaller number of electrons an atom has to borrow or to lend, the greater the activity.

The answer is B.

3 0

4 years ago

What is a compound?

hoa [83]

I think it’s C. A molecule that is formed from two or more different types of atoms.

4 0

3 years ago

What is the specific heat capacity of silver metal if 75.00 g of the metal absorbs 56.4J of heat and the temperature rises 35.0°

Eva8 [605]

Answer:

The specific heat capacity of silver metal is 0.021485 J/g°C.

Explanation:

The specific heat capacity of silver metal can be found out by using the formula,

Q = (mass) (ΔT) (Cp)

Given,

mass = 75 grams

Q = 56.4 j

ΔT = 35°C

Substituting values,

56.4 = (75)(35)(Cp)

Cp = 0.021485 J/g°C.

8 0

4 years ago

How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that t

Arturiano [62]

<u>Answer:</u> The amount of heat required is 775.7 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)$

Now, we calculate the amount of heat released or absorbed in all the processes.

<u>For process 1:</u>

$q_1=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_f$ = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

$q_1=248g\times 334J/g=84832J$

<u>For process 2:</u>

$q_2=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_2$ = amount of heat absorbed = ?

$C_{p,l}$ = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J$

<u>For process 3:</u>

$q_3=m\times L_v$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_v$ = latent heat of vaporization = $40.79kJ/mol\times \frac{1000}{18}=2266.1J/g$ (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)

Putting all the values in above equation, we get:

$q_3=248g\times 2260J/g=560480J$

<u>For process 4:</u>

$q_4=m\times C_{p,g}\times (T_{2}-T_{1})$

where,

$q_4$ = amount of heat absorbed = ?

$C_{p,g}$ = specific heat of steam = 1.99 J/g°C

m = mass of water = 248 g

$T_2$ = final temperature = $154^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_4=248g\times 1.99J/g^oC\times (154-100)^oC=26650.1J$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3+q_4$

$Q=[84832+103763.2+560480+26650.1]J=775,725.3J=775.7kJ$

Hence, the amount of heat required is 775.7 kJ

5 0

4 years ago

Is soda just opened a heterogeneous mixture?

grin007 [14]

Yes, because Google.

3 0

3 years ago