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nikitadnepr [17]
3 years ago
12

PLEAEEEE HELPPP

Chemistry
1 answer:
alexgriva [62]3 years ago
8 0
Speed is a scalar quantity( quantity having only magnitude). So it doesn't specifies direction. Hence option C.) 5Km/hr north is incorrect.
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What would be the effect on your calculated mas percent recovery of Cu if you did not completely react the Mg in the reaction mi
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A student prepares an iron standard solution in a 50.00 mL volumetric flask. The iron stock concentration is 0.0001974 M. A stud
Serhud [2]

Answer:

0.00011765 M

Explanation:

When a solution is prepared by dilution, the volumes and concentrations are related by:

C1*V1 = C2*V2

Where C1 is the concentration of the solution 1, V1 is the volume of the solution 1, C2 is the concentration of solution 2, and V2 is the volume of solution 2.

The stock solution is the solution 1, and the standard solution, the solution 2, so:

0.0001974*29.80 = C2*50.00

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4 0
3 years ago
QUESTION 7
Lera25 [3.4K]
Answer:
it’s quite simple.
The newton force in a aerodynamic space which could lead to a catastrophic entanglement will be added to the physical quantum molecular structure then subtracted by its opposite, inter dimensional indispensable aerobic activity, then finally multiplied by the sum which was subtracted by the addition to then divide the sum of it all with the solution of the multiplication which would equal to the answer.
4 0
2 years ago
What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?​
Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

    Volume of blood sample in uL = 2.15uL

Conversion           uL → mL

   micro- and milli-  are both prefixes of sub-units.

liter is a unit of volume of a substance.

       micro - is 10⁻⁶

       milli- is of the order 10⁻³

The problem is converting from micro to milli:

     if we multiply  10⁻⁶ by 10³ we would have our milli;

  1000uL = 1mL

  2.15uL :   2.15uL x \frac{1mL}{1000uL} = 2.15 x 10⁻³mL

learn more:

Volume brainly.com/question/5055270

#learnwithBrainly

4 0
3 years ago
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