Answer:
No, it's a physical reaction.
Explanation:
A chemical change produces new chemical compounds, but combining water and powder is just mixing the powder with the water. It's not a new compound.
I don't know how to really explain, sorry :)
Answer:
0.479 M or mol/L
Explanation:
So Molarity is moles/litres of solution...often written as M=mol/L
So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2
Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L
Now we just plug these guys in our molarity formula M=mol/L
M= 0.3035/0.634 = 0.479 M or mol/L
True....
Explanation
Cells use oxygen to release energy during photosynthesis/chlorophyll/cellular respiration. For photosynthesis, the starting materials are water and carbon dioxide, and the products are sugar and oxygen. For cellular respiration, the starting materials are oxygen and sugar, and the products are carbon dioxide and water.
The A horizon is a surface horizon that largely consists of minerals (sand<span>, </span>silt<span>, and </span>clay) and with appreciable amounts of organic matter. This horizon is predominantly the surface layer of many soils in grasslands and agricultural lands.<span>These </span>materials typically<span> accumulate through a process termed illuviation, wherein the </span>materials<span> gradually wash in from the overlying.</span>
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
