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amm1812
3 years ago
12

Some students do a lab experiment to find g the accelerartion due to gravity. The value they came up with experimentally was 11.

25m/s2. The actual value for g is 9.81 m/s2. What is their percent error?
a
3.9 %
b
14.7 %
c
1.44 %
d
-6 %
Physics
2 answers:
____ [38]3 years ago
4 0
\frac{11.25 - 9.81}{9.81} \times 100 = 14.7\%
The answer is (B)
slava [35]3 years ago
4 0

Answer:

b. 14.7%

Explanation:

Experimental value of gravitational acceleration “g” = 11.25 m/s^2

Actual value of gravitational acceleration “g” = 9.81 m/s^2

Percentage error = ((Experimental value – Actual value)/Actual value) × 100

Percentage error = ((11.25 – 9.81)/9.81) × 100                                          

Percentage error = 14.7%  

Hence, the percentage error in the value of “g” will be 14.7%.                                        

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With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s
vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

6.51=0.53 x velocity of softball

Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

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The percentage of fe in an iron ii sample was determined by redox titration with cr2o7 calculate the molarity of the standard k2
Archy [21]

The molarity is 0.018M and the percentage is 8.46%.

Net ionic reaction is

$\mathrm{Cr} 2 \mathrm{O}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H} 2 \mathrm{O}$

$1 \mathrm{~mol}$ of $\mathrm{Cr} 2 \mathrm{O} 7(-2)$ reacts with $6 \mathrm{~mol}$of $\mathrm{Fe}(2+)$ to form $6 \mathrm{~mol}$of $\mathrm{Fe}(3+)$ and $2 \mathrm{~mol}$ of $\mathrm{Cr}(3+)$

Find molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

Molarity = moles of$\mathrm{K} 2 \mathrm{Cr} 207 /$ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7(\mathrm{~L}$ )

Moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ = grams of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ molar mass of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

$=1.2275 \mathrm{~g} / 294.19 \mathrm{~mol}=0.0042 \mathrm{~g} / \mathrm{mol}$

Molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7=$ moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$/ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$(L)

$=0.0042 \mathrm{~g} / \mathrm{mol} / 0.250 \mathrm{~L}=0.017 \mathrm{M}$

Find molarity of $\mathrm{Fe}$ (II)

Molarity of $\mathrm{Fe}(\mathrm{II})=6 \times$molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 \times$volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$volume of $\mathrm{Fe}$ (II)

$=6 \times 0.017 \mathrm{M} \times 0.03598 \mathrm{~L} / 0.200 \mathrm{~L}$

$=0.018 \mathrm{M}$

Find moles of $\mathrm{Fe}(\mathrm{II})$

Moles of \mathrm{Fe}($ II) =molarity of $\mathrm{Fe}$ (II) $\times$ volume of $\mathrm{Fe}$ (II)

=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}

Find mass of Fe(II)

Mass of \mathrm{Fe}( II )= moles of \mathrm{Fe}( II) \times molar mass of $\mathrm{Fe}(\mathrm{II})$

$=0.004 \mathrm{~mol} \times 55.85 \mathrm{~g} / \mathrm{mol}$

$=0.205 \mathrm{~g}$

Find $\% \mathrm{Fe}(\mathrm{II})$ in unknown sample

$$\begin{aligned}\% \mathrm{Fe} &=(\text { mass of } \mathrm{Fe} / \text { weight of unknown sample }) \times 100 \\&=(0.205 \mathrm{~g} / 2.4234 \mathrm{~g}) \times 100 \\&=8.46 \%\end{aligned}$$

  • Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution.
  • concentration is the amount of a substance per defined space. Concentration usually is expressed in terms of mass per unit volume.

To know more about  MOLARITY    visit : brainly.com/question/8732513

#SPJ4

7 0
2 years ago
WhAt is this question??????????
valina [46]

Answer:

Period = 7

Group = Actinides group

Family = Actinides Family

Explanation:

uranium found in seventh row or seventh period of the periodic table.

uranium is the member of Actinoides group it is also called Actinide group, And it has element whose atomic is number is 89 to 103. The atomic number of the uranium is 92. So, the uranium element is belongs to Actinide group

Uranium found in actinide family of periodic table, and actinides family has element whose atomic number is greater than or equal to 89 and less than or equal to 103.

8 0
3 years ago
NO LINKS OR I WILL REPORT YOU b/t/w i\d\k if this is physics or not
MrMuchimi

Above information helps you bro.

4 0
3 years ago
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