Answer:
see attached
Explanation:
Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.
Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.
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Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.
On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.
Answer:
1) The atomic number of this element is 74, because number of protons = atomic number
2) Mass number of this element is 184
Ar=Z+N=74+110=184
3) It isn't a new element. An element with such atomic and mass numbers exists
4) It is Wolfram. Symbol: W
5) Solid
Sorry if I made mistakes. I started learning English recently.
The solution for this problem is:
1. number of mole / kg of vitamin K = (4.43 °C) / (40.0 °C/m) = 0.11075 m *molality freezing point depression constant for camphor is 37.7 C Kg/mo
2. (0.11075 mol/kg) x (0.0100 kg) = 0.0011075 mol
3. (0.500 g) / (0.0011075 mol) = 451 g/mol
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Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane