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Elza [17]
3 years ago
14

Observation made when lead oxide passes through hydrogen gas​

Chemistry
1 answer:
CaHeK987 [17]3 years ago
7 0

Answer:

a gray substance and colourless liquid

Explanation:

lead oxide is reduced to lead while hydrogen is oxidized to water

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What volume of 0.128 M HCl is needed to neutralize 2.87 G of Mg(OH)2?
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V= .768L= 76.8ml HCL
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4 years ago
A student calculates the volume of a graduated cylinder to be 43.26 ml. The actual volume is 42.32 ml. What is the percent error
Tpy6a [65]

Answer:

2.2%

Explanation:

Percentage error,

You apply the formula,

[(Estimated value - Actual value)/Actual value] × 100%

; [(43.26 - 42.32)/42.32] × 100

; (0.94/42.32) × 100

; 0.022 × 100

Percent error = 2.2%

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4 years ago
Characteristics of bipolar disorder typically include
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C. Difficulty in concentrating. 
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3 years ago
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The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu
MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation-     k=Ae^{\frac{-E_{a}}{RT}}    , where k is rate constant, A is pre-exponential factor, E_{a} is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

                                   \frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}

Here \frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005} , T = 298 K , R = 8.314 J/(mol.K) and E_{a}^{ala-pro}=60kJ/mol

So, \frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}

   \Rightarrow E_{a}^{phe-pro}=65705J/mol=66kJ/mol (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0
3 years ago
Calculate the relative formula mass for KMnO4
GarryVolchara [31]

Explanation:

please mark me as brainlest

4 0
3 years ago
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