You should of might do this first take a picture of the problem to make it easy for us to understand just saying for the furture
Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams
i think..
potassium hydroxide..
The answer to the question is a
C) Qc < Kc, the reaction proceeds from left to right to reach equilibrium
<h3>Further explanation</h3>
Given
K = 50.2 at 445°C
[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC
Reaction
H2(g) + I2(g) ⇔2HI(g)
Required
Qc
Solution
Qc for the reaction
![\tt Qc=\dfrac{[HI]^2}{[H_2][I_2]}\\\\Qc=\dfrac{(1.75.10^{-3})^2}{1.75.10^{-3}\times 1.7\times 10^{-3}}=1](https://tex.z-dn.net/?f=%5Ctt%20Qc%3D%5Cdfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D%5C%5C%5C%5CQc%3D%5Cdfrac%7B%281.75.10%5E%7B-3%7D%29%5E2%7D%7B1.75.10%5E%7B-3%7D%5Ctimes%201.7%5Ctimes%2010%5E%7B-3%7D%7D%3D1)
Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)
