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sergij07 [2.7K]
1 year ago
13

Need answer asap 4 both

Chemistry
1 answer:
marishachu [46]1 year ago
7 0

The persistence of vision is regarded as an optical illusion but it is actually part of human physiology.

<h3>What is persistence of vision?</h3>

The term persistence of vision is the idea that an object is able to remain is sight for a long time even when the object has been removed. This is often used by performers to create a remarkable impression on the audience. The persistence of vision is regarded as an optical illusion but it is actually part of human physiology.

The way that a performance can be designed to make the audience to know that this is the most important scene is called follow through.

Learn more about optical illusion:brainly.com/question/28179807

#SPJ1

You might be interested in
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
M. element<br>cenabrecht<br>0.5 mele of an​
drek231 [11]

Answer:

Number of moles (n)=

molecular weight

weight

Weight=n×Molecular weight

=0.5×14

Mass=7g

5 0
3 years ago
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
Liters of a 2.40 M
kotykmax [81]

Answer:

1.42 L

Explanation:

Step 1:

The following data were obtained from the question :

Molarity of KBr = 2.40 M

Mole of KBr = 3.40 moles

Volume of solution =?

Step 2:

Determination of the volume of the solution.

Molarity of solution is simply the mole of the solute per unit volume the of solution. It is given as :

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 3.4/2.4

Volume = 1.42 L

Therefore, the volume of the solution is 1.42 L

6 0
3 years ago
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