Answer:
Explanation:
Hydrogen has 1 proton and one nuetron
Oxygen has 8 protons nad 8 neutrons
Answer:
1. C
2.B
Explanation:
Answer:
3,8×10⁻⁵ mol/L of potassium permanganate solution
Explanation:
To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.
3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.
100 mL × ( 1L / 1000 mL) = 0,100 L
Thus, concentration in mol/L is:
3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution
I hope it helps!
1. Synthesis reaction : there is only 1 product formed from 2 or more reactant
E.g:
2. Decomposition : reaction that occurs in presence of UV light and only 1 reactant that decomposes into 2 or more products.
E.g:
3. Single displacement :reaction that occurs when 1 reactant displaces other reactant from its compound:
E.g:
4. Double displacement :reaction that occurs when both reactant displaces each other.
E.g :
As the temperature of a liquid or stable will increase its vapor strain additionally will increase. Conversely, vapor strain decreases because the temperature decreases.
The better the vapor strain of a substance, the extra the awareness of the compound withinside the gaseous section and the extra the quantity of vaporization
. Liquids range substantially of their vapor pressures. substance with a excessive vapor strain at everyday temperatures is regularly called volatile. The strain exhibited through vapor gift above a liquid floor is referred to as vapor strain. As the temperature of a liquid will increase, the kinetic strength of its molecules additionally will increase.
Read more about vapour:
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Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g
periodic poperties of elements in groups
and periods
atomic radius
ionization potential