Answer:
0.64 g of S
Solution:
The balance chemical equation is as follow,
2 Cu + S ----> Cu₂S
According to equation,
127 g (2 mole) Cu produces = 159 g (1 mole) of Cu₂S
So,
2.54 g Cu will produce = X g of Cu₂S
Solving for X,
X = (2.54 g * 159 g) / 127 g
X = 3.18 g of Cu₂S
Now, it is confirmed that the reaction is 100% ideal. Therefore,
As,
127 g (2 mole) Cu required = 32 g (1 mole) of S
So,
2.54 g Cu will require = X g of S
Solving for X,
X = (2.54 g * 32 g) / 127 g
X = 0.64 g of S
<u>step</u><u> </u><u>by</u><u> </u><u>step</u>
Na(sodium)=2.8.1
Cl (Chlorine)=2.8.7. sodium will give the chlorine the 1 valence electron to become stable ions.
<u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
<u>p</u><u>o</u><u>t</u><u>t</u><u>a</u><u>s</u><u>i</u><u>u</u><u>m</u><u>. </u><u>(</u><u>2</u><u>.</u><u>8</u><u>.</u><u>8</u><u>.</u><u>1</u><u>)</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>F</u><u>l</u><u>o</u><u>u</u><u>r</u><u>i</u><u>n</u><u>e</u><u>(</u><u>2</u><u>.</u><u>7</u><u>)</u>
Here are a few examples :)
iodine (I2)
naphthalene
aresenic (As)
ferrocene
water (H2O)
carbon dioxide (CO2)
Hope this helps :)
x = 1.01
Explanation:
Given equation:
y = 1.2345x – 0.6789
y = 0.570
Problem:
Solving for x
The variables in this equation are y and x
They can take up any value since they are variables.
Since we have been given y = 0.570
y = 1.2345x – 0.6789
To solve for x, we simply substitute for y in the equation;
since y = 0.570
0.57 = 1.2345x – 0.6789
add 0.6789 to both sides;
0.57 + 0.6789 = 1.2345x – 0.6789 + 0.6789
1.2489 = 1.2345x
Divide both sides by 1.2345
x = 1.01
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Answer:
The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present
Explanation:
In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.
We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.