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tensa zangetsu [6.8K]
2 years ago
13

Can someone help me please

Chemistry
1 answer:
almond37 [142]2 years ago
5 0

Answer:D

Explanation:

  1. Use the vertex form, y=a(x−h)2+ky=a(x-h)2+k, to determine the values of aa, hh, and kk.a=14a=14h=6h=6k=1k=1Find the vertex (h,k)(h,k).(6,1)(6,1)Find pp, the distance from the vertex to the focus.
  2. Find the distance from the vertex to a focus of the parabola by using the following formula.14a14aSubstitute the value of aa into the formula.14⋅1414⋅14Simplify.
  3. Combine 4 and 1/4 .1 /4/4
  4. Simplify by dividing numbers
  5. Divide 4 by 44.1/1Divide 1 by 1/1.
  6. Y=0
  7. The Answer is D
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Answer:

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5 0
3 years ago
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 405 mL .
bezimeni [28]
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7 0
3 years ago
A 96.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 48.0 ml of koh at 25 ∘c.
PilotLPTM [1.2K]
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8 0
3 years ago
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If 5 moles of P4 reacted with 22 moles Cl2 according to the above reaction, determine:
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Solution:

P_4+6Cl_2\rightarrow 4PCl_3

a) By stoichoimetry If 6 moles of Cl_2 gives 4 mole of PCl_3 , then 22 moles of Cl_2 will give: \frac{4}{6}\times 22=\frac{44}{3}=14.66 moles of PCl_3

b) If 6 moles of Cl_2 reacts with one mole of P_4 , then 22 moles of Cl_2 will react : \frac{1}{6}\times 22=\frac{11}{3}=3.66 moles of P_4

Moles of P_4  left after the reaction = 5-3.66=\frac{4}{3}=1.34 moles.

c) 0 Moles of Cl_2 , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.



7 0
3 years ago
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