Answer:
the heat from the universe during the big bang
a.k.a.
ITS A
Explanation:
the universe is still expanding, so what is gonna be used to create that new matter that's expanding the universe? energy and heat from the beginning of the universe! the universe started off as this point that was probably smaller than a quantum foam and infinite heat and started expanding from there
Answer: One major strategy by which plants adapt to temperature change is to decrease the degree of unsaturation of membrane lipids under high temperature and increase it under low temperature. ... This pattern of changes in membrane lipids was distinct from that occurring during slow alterations in temperature.
Explanation: I found this answer from goggle
A mixture of Cu2 and CuO of mass 8.828g is reduced to copper metal with hydrogen:
Cu2O + H2 --> 2Cu + H2O
CuO + H2 --> Cu + H2O
If the mass of pure copper isolated was 7.214g, determine the percent by mass of CuO in the original sample
Let x = grams of CuO in the original sample.
y = grams of Cu2O in the original sample.
Eq. #1 x + y = 8.828 grams
Molar mass of CuO = 63.5 + 16 = 79.5 grams
Moles of CuO = x ÷ 79.5
Molar mass of Cu2O = 63.546 + 32 = 95.5 grams
Moles of Cu2O = y ÷ 95.5
According to the 2nd balanced equation, CuO + H2 --> Cu + H2O ,
1 mole of CuO produces 1 mole of Cu.
So, x ÷ 79.5 moles of CuO will produce x ÷ 79.5 moles of Cu
According to the 1st balanced equation, Cu2O + H2 --> 2Cu + H2O,
1 mole of Cu2O produces 2 moles of Cu
So, (y ÷ 95.5) moles of Cu2O will produce 2 * (y ÷ 95.5) moles of Cu
Since, the mass of pure copper isolated was 7.214 grams
Moles of Cu = (7.214 ÷ 63.5)
Moles of Cu from Cu2O + moles of Cu from CuO = total moles of Cu!!
2 * (y ÷ 95.5) + (x ÷ 79.5) = (7.214 ÷ 63.5)
Multiply by both sides by 95.5 * 79.5 * 63.5 to get rid of denominators
(2 * 79.5 * 63.5) y + (95.5 * 63.5) x = (7.214 * 95.5 * 79.5)
10,096.5 y + 6,064.25 x = 36,418.0755
Divide both sides by 6,064.25
x + 1.665 y = 6
Eq.#2 x = 6 – 1.665 y
Eq. #1 x + y = 8.828
x = 8.828 – y
8.828 – y = 6 – 1.665 y
0.665 y = 2.828
y = 4.25 grams of Cu2O
x = 8.828 – 4.25 = 4.58 grams of CuO
% CuO = (4.58 ÷ 8.828) * 100 = 51.88% CuO
0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
<h3>What is dilution?</h3>
Dilution is a process of making a solution of lower concentration from a solution of higher concentration by the addition of solvent to a given volume of the solution of higher concentration.
Dilution of solutions is done using the dilution formula in order to determine the given volume of diluent or stock solution required. The dilution formula is given below:
where:
- C1 = Initial concentration of enzyme
- C2 = Final concentration of enzyme
- V1 = Initial volume
- V2 = Final volume
For the enzyme dilution;
C1 = 1 mg/mL
C2 = 1/50 mg/ml = 0.02 mg/ml
V= ?
V2 = 5 ml
V1 = C2V2/C1
V1 = 0.02 * 5/1 = 0.1 mL
Therefore, 0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
Learn more about dilution at: brainly.com/question/24881505
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Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent composition of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
