Answer:
(1)There are 1.5 moles of water in a 27 gram sample of water. The molar mass of water is 18.02 gmol g m o l .(2)
AnswersChemistryGCSEArticle
What is the mass (g) of 0.25mols of NaCl?
What you need for these equations are a calculator, periodic table and the following equation:
Mass (g) = Mr x Moles (important equation to remember)
In this case we already know the moles as it's in the question, 0.25 moles.
to find the Mr, you need to look at your periodic table. Find the relative atomic mass of Na and Cl and add the two numbers together.
Na = 22.99
Cl = 35.45
NaCl = 58.4
Now just put all of the numbers into the equation.
0.25 x 58.4 = 14.6g
You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.
Explanation:
To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:
2 Mg (s) + O₂ (g) → 2 MgO + heat
CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat
However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)
So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.
we have used the following notations:
(s) - solid
(g) - gas
(l) - liquid
Learn more about:
combustion reactions
brainly.com/question/13824679
#learnwithBrainly
A force of attraction that
holds atom together
When atoms react they form a
chemical bond which is defined as a force of attraction that holds atom
together. A force of attraction is defined as a kind of force that draws two or
more objects together regardless of distance. There are two major categories of
forces of attraction, one is intramolecular and intermolecular. Intramolecular forces
is the presence of forces in atoms internally. While intermolecular is the
force by which the force that is existent in two or more elements.
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
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