M = Sr (strontium)
<em>Step 1.</em> Calculate the <em>moles of CO_2</em>.
Moles of CO_2 = 0.395 g CO_2 × (1 mol CO_2/44.01 g CO_2)
= 0.008 975 mol CO_2
<em>Step 2</em>. Calculate the <em>moles of MCO_3</em>.
Moles of MCO_3 = 0.008 975 mol CO_2 × (1 mol MCO_3/1 mol CO_2)
= 0.008 975 mol MCO_3
<em>Step 3</em>. Calculate the molar mass of <em>MCO_3</em>
MM = grams/moles = 1.324 g/0.008 75 mol = 147.5 g/mol
<em>Step 4</em>. Calculate the <em>atomic mass of M</em>
M_r = <em>x</em> + 12.01 + 3×16.00 = <em>x</em> + 60.01 = 147.5
<em>x</em> = 147.5 – 60.01 = 87.5
<em>Step 5. Identify M</em>.
The element with the closest atomic mass is Sr (A_r = 87.6).
∴ M = Sr and the compound is SrCO_3.
It’s [Xe]7s^2 5f^14 (first one) hopefully i could help!!
Answer:
second order
Explanation:
units of reaction and their order.
Zero order --> M^1 s^-1 = M/s
First order --> M^0 s^-1 = 1/s
Second order --> M^-1 s^-1 = L/mol s
In the question rate constant k = 4.65 L mol-1s-1. = 4.65 L/mol s
Hence, the reaction is a second order reaction
Answer:
The stationary phase in chromatography experiment is paper.
Explanation:
In chromatography experiment, the stationary phase is defined as the fixed substance that is necessary to start chromatography. In our case, this fixed substance is paper, so that makes paper our stationary phase.
Hope this helps :)
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
