Are the particles in a solid in constant motion, or are they "frozen in place?"

The rate constant for a first-order reaction is 0.54 s-1. What is the half-life of this reaction if the initial concentration is

Diano4ka-milaya [45]

Answer:

The half life time of first order reaction is 1.3 sec

Explanation:

Given:

First order rate constant $K = 0.54$ $M^{-1} s^{-1}$

Initial concentration $= 0.54$ M

From the formula of first order half life time,

$K = \frac{0.693}{t_{\frac{1}{2} } }$

So half life time is given by,

${t_{\frac{1}{y2} } } = \frac{0.693}{0.54}$

${t_{\frac{1}{y2} } } = 1.3$ sec

Therefore, the half life time of first order reaction is 1.3 sec.

4 0

3 years ago

How many significant figures are present in 2.020 x 10-25

garik1379 [7]

Answer:4 and 3

Explanation:just took the test

3 0

3 years ago

El sol es la estrella más grande

Misha Larkins [42]

Answer:

Schb

Explanation:

5 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

Control of Blood pH by respiratory rate.

nata0808 [166]

Answer:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

Explanation:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

$CO_{2} +H_{2} O$ ⇄ $H^+ + HCO^-_{3}$

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in $H^+$ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

6 0

3 years ago