Answer:
MoClBr₂
Explanation:
First we calculate the mass of bromine in the compound:
- 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g
Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:
- 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
- 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
- 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br
Now we <u>divide those numbers of moles by the lowest number among them</u>:
- 0.9594 mol Mo / 0.9594 = 1
- 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
- 2.150 mol Br / 0.9594 = 2.24 ≅ 2
Meaning the empirical formula is MoClBr₂.
<span>We have ground strate configurations of electrons,if electrons are filled in order of increasing energy. When there are electrons are in higher orbitals, we have an atom in an excited state.
B, and C are excited states.
In B, 2 electrons can fit in the 4s orbital, and that should fill fully before the 4p orbitals.
In C, the same is true for 5s and 5p
In D, this is not an excited state because 4s fills before 3d</span>
Its the answer b, because oil has a great density, therefore, when you mix them together the oil will be below and the above of it.
Answer:
= 15.51 mL
Explanation:
Here's is the reaction:
2HgO(s) ⇒ 2 Hg(s)+O₂(g)
In this reaction 2mol HgO = 1mol O₂
The molecular weight of HgO = 216.59g
so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO
= 0.0138511 molHgO
The amount of Oxygen follows:
0.0138511 molHgOx1/2= 0.00692555 mol O₂
Now, volume of 1 any gas = 22400mL
so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂
= 15.513232mL O₂
Answer:
7200 kPa
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas
Make P' the subject of the equation
P' = PVT'/V'T.............. Equation 2
Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K
Substitute these values into equation 2
P' = (900×8×600)/(2×300)
P' = 7200 kPa