Idk i just need to ask a question
Sodium, Atomic mass: 22.989769 g
You can see in a periodic table
<h3>
Answer:</h3>
0.10 L
<h3>
Explanation:</h3>
The concentration of glucose is given as 180 g/L
The mass of glucose is 18 g
- Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
- When calculating molarity on the other hand, we divide number of moles of the solute by the volume of the solution.
- Concentration in g/L = Mass of solute ÷ Volume
Rearranging the formula,
Volume = Mass of the solute ÷ concentration
= 18 g ÷ 180 g/L
= 0.10 L
Therefore, volume of water is 0.10 L
According to an article dated back in February 8, 1992 which is entitled, “Science: Stardust is made of diamonds” on a website called newscientist (https://www.newscientist.com/article/mg13318073-000-science-stardust-is-made-of-diamonds/), American astronomers believed that diamonds are made in supernova explosions. It was said that the diamonds were the foundation of uncommon combinations of isotopes found in some meteorites. Donald Clayton of Clemson University in South Carolina suggested that the weightiest isotopes were more common in meteorites for the reason that the rare gases shaped in the neutron-rich outcome of a supernova explosion. Clayton also said, “the observed mixture of isotopes could have been produced only during the collapse of a massive star to form a neutron star”. This happens in a Type II explosion, for example the Supernova 1987A in the Large Magellanic Cloud. And rare gases like xenon become stuck in both weighty and light isotopes after the ejected gas from such a supernova cools down enough to create dust. The existence of the diamonds with these unusual gases in meteorites infers an alike source. Some of the carbon in the supernova fragments produces ordinary graphite dust, whereas some produces diamond dust. Considerable amount of stardust may be made of diamonds, if Clayton was not mistaken.
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
