The radioactive isotope I-131 is used in

Balance this chemical formula Fe + Cl2 = FeCl3

san4es73 [151]

First you find out how much each element has

Fe=1 * 2 Fe=1
Cl=2 *3 Cl=3 *2

now we multiply each so we can balance each side.

So now we get our balanced equation
2 Fe + 3 Cl2 = 2 FeCl3

5 0

3 years ago

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What is the formula for germicide that is sufficient to kill blood-borne pathogens?

Colt1911 [192]

It is 1 ounce. Bloodborne Pathogens can be transmitted when blood or body liquid from a tainted individual enters someone else's body by means of needle-sticks, human chomps, cuts, scraped areas, or through mucous films. Likewise, semen, vaginal discharges and salivation in dental methods are considered conceivably tainted body liquids.

7 0

3 years ago

Identify the cfc with the shortest atmospheric lifetime

goldenfox [79]

CFC-11 is typically the CFC with the shortest atmospheric lifetime.

5 0

3 years ago

Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe

Viktor [21]

There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.

7 0

3 years ago

If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th

qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

$K_c=4.74\times 10^4$

Concentration at equilibrium:

$[COBr_2]=1.58\times 10^{-6}M$

$[Co]=2.78\times 10^{-3}M$

$[Br_2]=2.51\times 10^{-5}M$

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][Br_2]}{[COBr_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}$

The given equilibrium constant value is, $K_c=4.74\times 10^4$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q$ that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0

3 years ago