On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.
Answers
Answer:
Average atomic mass = 46.91466 amu
Explanation:
Step 1: Data given
Isotopes of titanium
46Ti = 70.900% ⇒ 45.95263 amu
48Ti = 10.000 % ⇒ 47.94795 amu
50Ti = 19.100 % ⇒ 49.94479 amu
Step 2: Calculate the average atomic mass of titanium
Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479
Average atomic mass = 46.91466 amu
Related Questions
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Write Lewis structural formulas for the following molecules: NH3 SO2 CH3OH HNO2 N2 CH2O
At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?
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Answers
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:
Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:
I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:
Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
Answers
Answer:
Moles of hydrogen formed = 3.5 moles
Explanation:
Given that:-
Moles of aluminium= 4.0 mol
Moles of hydrogen bromide = 7.0 mol
According to the reaction:-
2 moles of aluminum react with 6 moles of hydrogen bromide
1 mole of aluminum react with 6/2 moles of hydrogen bromide
4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide
Moles of hydrogen bromide = 12 moles
Available moles of hydrogen bromide = 7.0 moles
Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)
The formation of the product is governed by the limiting reagent. So,
6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen
1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen
7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen
Moles of hydrogen formed = 3.5 moles
Answer:
3.5 mol H2, HBr (limiting reactant)
Explanation:
4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2
7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2
Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.
Answers
c) potassium, K(s)
b) manganese, Mn(s)
d) boron, B(s)
Answers
Answer:
The molar mass of:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Explanation:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Answers
The internal energy : 1310.43 kJ/mol
Further explanation
Internal energy (ΔE) can be formulated for Calorimeter :
C= the heat capacity of the calorimeter
Δt=36.2-25.68=10.52°C
- mol butane(MW=58,12 g/mol)
- the internal energy (ΔE) in KJ/mol
Final answer:
The change in internal energy when 2.67 grams of butane is combusted in a bomb calorimeter, given a temperature increase from 25.68 C to 36.2C and a heat capacity of 5.73 kJ/C for the calorimeter, is approximately 1308 kJ/mol.
Explanation:
To solve the problem of calculating the changes in internal energy when 2.67 grams of butane (C4H10) is combusted in a bomb calorimeter, it is necessary to understand calorimeter's heat capacity and how a bomb calorimeter works.
The first step will be to calculate the change in temperature which here is the final temperature subtracted from the initial temperature: 36.2 C - 25.68 C = 10.52 C.
Then, we multiply this temperature change by the heat capacity of the calorimeter to find the total heat produced by the reaction in kJ: 10.52 C * 5.73 kJ/C = 60.18 kJ.
The final step is to convert grams of butane to moles, because we are asked to find the energy change in kJ/mol. The molar mass of butane (C4H10) is approximately 58.12 g/mol. So we have approximately 2.67 g / 58.12 g/mol = 0.046 mol.
Finally, we divide the heat produced by the number of moles to get the energy change per mole of butane: 60.18 kJ / 0.046 mol = approximately 1308 kJ/mol.
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B) Respiration O2 photosynthesis CO2
C) Photosynthesis H2 Respiration O2
D) Respiration CO2 photosynthesis H2
Answers
the answer is A:
Photosynthesis release O2 (oxygen)
then animals use oxygen for respiration and release Co2
Final answer:
The correct answer is A) Photosynthesis O2 Respiration CO2. Photosynthesis converts CO2 to O2, which is then used in respiration to be converted back into CO2. Both processes together create a continuous cycle.
Explanation:
The correct diagram that represents the cycling of gases between photosynthesis and respiration is A) Photosynthesis O2 Respiration CO2. Through the process of photosynthesis, plants convert carbon dioxide (CO2) and sunlight into oxygen (O2) and glucose. Organisms, including the plants themselves, then use that oxygen for respiration, during which they convert the oxygen back into carbon dioxide. The glucose is used for energy. This continuous cycling plays a critical role in life on Earth.
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