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zhuklara [117]
4 years ago
8

Read the graph to the nearest tenth for both pH and volume of calcium hydroxide. 25.0ml of a 1.70x10-4M solution of Hydrochloric

acid was titrated with Calcium hydroxide. The above graph was generated when the Hydrochloric acid was titrated with Calcium hydroxide. Determine the concentration (in M) of the Calcium hydroxide. What is the coefficient of the scientific notation answer for the concentration of Calcium Hydroxide.
Determine the percent error if the known concentration of calcium hydroxide is 6.30x10-4M. (Do not put your answer in scientific notation).

Chemistry
1 answer:
-BARSIC- [3]4 years ago
4 0

Answer:

Concentration of base obtained from the graph provided = (1.93 × 10⁻⁴) M

The coefficient of the scientific notation = 1.93

Percent error = 69.36%

Explanation:

The balanced stoichiometric chemical reaction between HCl and Ca(OH)₂ is given as

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Using the relation for neutralization titration

(Ca × Va)/(Cb × Vb) = (na/nb)

Ca = Concentration of acid = (1.70 × 10⁻⁴) M

Va = Volume of acid = 25.0 mL

Cb = Concentration of base = ?

Vb = 11.0 mL (from the graph attached, the reaction reaches endpoint at 11.0 mL volume of base.

na = coefficient of acid in the balanced equation = 2

nb = coefficient of base in the balanced equation = 1

(1.70 × 10⁻⁴ × 25)/(Cb × 11) = (2/1)

(Cb × 11) = (0.00017 × 25/2) = 0.002125

Cb = (0.002125/11) = 0.0001931818

= (1.93 × 10⁻⁴) M

Percent error = 100% × [(difference between wrong value and true value) ÷ (true value)]

Difference between wrong value and true value = (6.30 × 10⁻⁴) - (1.93 × 10⁻⁴) = (4.37 × 10⁻⁴)

True value = (6.30 × 10⁻⁴)

Percent error

= 100% × [(4.37 × 10⁻⁴) ÷ (6.30 × 10⁻⁴)]

= 69.36%

Hope this Helps!!!

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atroni [7]

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

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5 0
4 years ago
Can someone please please help me out ?
GarryVolchara [31]

Answer:

4.823 x 10^-19 J

Explanation:

Energy is calculated by E = hv where h - Planck's constant in joule.s

v - frequency.

in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s

-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec

now we can simply use Planck's constant in E=hv =

(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.

6 0
3 years ago
A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r
svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

8 0
3 years ago
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