The electronic configuration of Chlorine is 1s2 2s2 2p6 3s2 3p5.
There are three energy levels in chlorine
First energy level is n=1 has 1s2 so total 2 electrons
Second energy level is n=2, 2s2 2p6 so total 2+6= 8 electrons
Third has 3s2 3p5 electrons just 7 total... p can contain 6 electrons but only 5 are present. So the third level has lesser number of electrons than that can be filled
Answer:
mCO2= 49.6932 kgCO2
Explanation:
Hello! Let's solve this!
First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O
We see that each mole of C3H8 (propane) we get 3 moles of CO2
From the propane volume we can obtain the grams of propane used.
molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane
mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2
mCO2= 49.6932 kgCO2
Answer:
<h2>Dog's mitochondria lack the transport protein that transport pyruvate ( end product of glycolysis) across the outer mitochondrial membrane
.</h2>
Explanation:
1. As given here that dog's mitochondria can use only fatty acids and also amino acids for their respiration, and as compared to others, Dong's cell produce more lactate then normal, this indicate that his mitochondrial membrane is different then others.
2. The aerobic phases of cellular respiration in eukaryotes occur within mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.
3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial membrane.
Answer:
The answer to your question is: 234.7 cans
Explanation:
data
caffeine concentration = 3.55 mg/oz
10.0 g of caffeine is lethal
there are 12 oz of caffeine in a can
Then
3.55 mg ----------------- 1 oz
x mg -----------------12 oz (in a can)
x = 42.6 mg of caffeine in a can
Convert it to grams 42,6 mg = 0.0426 g of caffeine in a can
Finally
0.0426 g of caffeine ------------------ 1 can
10 g of caffeine ----------------- x
x = 10 x 1/0.0436 = 234.7 cans
Answer:
CaCO₃(s) => CaO(s) + CO₂(g) ... GpIIA Decomp
Explanation:
Metallic Carbonates decompose into a metallic oxide and carbon dioxide.
Examples:
Na₂CO₃(s) => Na₂O(s) + CO₂(g) ... GpIA Decomp
MgCO₃(s) => MgO(s) + CO₂(g) ... GpIIA Decomp
