Answer:
The adhesive is made up of two different layers.
The lower layer is made up of vertically aligned nanotubes.
The upper layer is made up of curly, spaghetti-like, nanotubes.
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Answer:
the proton and the neutron
Explanation:
mass of The proton is approximately 1.6726 × 10^-27 Kg
mass of the electron is approximately 9.109 × 10^-31 Kg
mass of the neutron is approximately 1.6749 × 10^-27 Kg
Here we see that mass of proton and neutron is approximately similar.
_____________________________________________
Lets see what is the mass of these particle IN atm unit
mass of these particles can also be described in form of amu(atomic mass unit)
1 atm is the mass defined as 1/12 th of mass of the carbon-12(c-12) atom.
In terms of unit atm
mass of proton = 1 amu
mass of electron = 5.45 × 10−4 amu
mass of neutron = 1 amu
hence we can say that mass of proton = mass of neutron.
mass of atom is sum of mass of all the neutron and proton in the atom.
Hence, mass of neturon and atom cannot be equal.
This makes option four the neutron and the atom , the wrong choice.
correct answer is the proton and the neutron
therefore, An atom contains one proton , one electron and one neutron then, mass of proton is similar in mass of neutron.
Answer:
1. the group number of sodium is 1 and it is a metal
2. the group number of helium is 18 and it is a nonmetal
3. the group number of iodine is 17 and it is a nonmetal
4.the group number of calcium is 2 and it is a metal
5. lithium has similar properties to potassium
6. calcium has similar properties to magnesium
7. neon has similar properties to xenon
8. Iodine has similar properties to chlorine
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
