Answer:
H2O
Explanation:
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The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.
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Yes the water boils but it doesn't become water gas it becomes just gas also known as evaporation
<span>361.4 pm is the length of the edge of the unit cell.
First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight.
Atomic weight copper = 63.546
Now divide the mass by the atomic weight, getting
8.94 g / 63.546 g/mol = 0.140685488 mol
And multiply by Avogadro's number to get the number of atoms:
0.140685488 * 6.022140857x10^23 = 8.472278233x10^22
Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume.
8.472278233x10^22 / 4 = 2.118069558x10^22
Now calculate the volume each unit cell occupies.
1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3
Let's get the cube root to get the length of an edge.
(4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm
Now let's convert from cm to pm.
3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm
Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
