Answer:
Steak is abotic living organisms are abotic and steak was abiotic
Explanation:
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
The final volume of the unknown gas is 787.5 ml
Given:
volume of unknown gas = 450.0 mL
initial pressure of unknown gas = 0.07 atm
final pressure of unknown gas = 0.04 atm
To Find:
final volume of the unknown gas
Solution:
Substituting these values into Boyle’s law, we get
P1V1 = P2V2
(0.07)(450) = (0.04)V2
V2 = (0.07)(450)/(0.04)
V2 = 787.5
So, final volume of the unknown gas is 787.5 ml
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Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
<em>Moles HCl added:</em>
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
<em>Moles NaOH to titrate the excess:</em>
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
<em>Moles of acid that were neutralized:</em>
0.0105 moles - 0.0010 moles =
<h3>0.0095 moles of acid were neutralized by the antiacid</h3>
They are the same type of concentration
