A car is traveling 78.9 kilometers per hour. What is its speed in miles per minute? One ilometer = 0.62miles

How to balance equations for reduction/oxidation reaction?

KATRIN_1 [288]

The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction, which requires that species exhibits changing oxidation statesduring the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components and is often separated into independent two hypothetical half-reactions to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:

Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)(1)(1)Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)

The first step in determining whether the reaction is a redox reaction is to splitting the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:

Cu(s)→Cu2+(aq)(2a)(2a)Cu(s)→Cu2+(aq)

The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in Cu to +2 in Cu2+. Now consider the silver atoms

2Ag+(aq)→2Ag(s)(2b)(2b)2Ag+(aq)→2Ag(s)

In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an element on its own. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.

Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.

6 0

3 years ago

Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give y

Andreyy89

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

$A = \lambda N$ (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

$N = \frac{N_{A}*m}{M}$ (2)

Where:

$N_{A}$ : is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol

Now, the decay constant is given by:

$\lambda = \frac{ln(2)}{t_{1/2}}$ (3)

Where:

${t_{1/2}}$ : is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

$A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s$