Need help with 4 through 7

Which of the following processes does not facilitate containment?

Marta_Voda [28]

Enrichment of uranium-235 is done. Hence, option A is correct.

<h3>What is the meaning of containment?</h3>

The act, process, or means of keeping something within limits the containment of health costs.

Enrichment is a process of increasing the proportion of fissile isotope found in uranium ore (represented by the symbol 'U') to make it usable as nuclear fuel or the compressed, explosive core of nuclear weapons.

Hence, option A is correct.

Learn more about containment here:

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7 0

2 years ago

Lana balanced an equation so that the result was 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr. Which most likely represents the starting e

seropon [69]

Answer:

C2H3Br + O2 → CO2 + H2O + HBr

Explanation:

The term balancing of chemical reaction equation has a unique meaning in chemistry. What it actually means is to ensure that the number of atoms of each element on the left hand side of reaction equation becomes equal to the number of atoms of the same element on the right hand side of the reaction equation.

When we look at the equation; C2H3Br + O2 → CO2 + H2O + HBr, the number of atoms of each element on the left and right hand sides of the given equation are not the same hence the equation is unbalanced.

If we look at the equation; 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr, the number of atoms of each element on both sides of the reaction equation are now equal, thus the later equation is the balanced version of the former.

6 0

3 years ago

Okay so this is half math half chemistry...

AleksandrR [38]

So you need to find the volume in L? If so:
Convert the mass of Lithium Bromide into moles by dividing the 100 grams by the molar mass of LiBr, taken from the periodic table
In a solution, moles = (concentration in mole/L) x (volume in L)
We know the moles, we have the concentration in mole/L, now find the volume in L, and you should get 0.288. Plz do the math and check for yourself

3 0

3 years ago

The equilibrium constant for the reaction

Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

Answer: The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$ .......(1)

We are given:

Volume of the container = 1.0 L

Value of $K_c$ = 2.0

Molarity of the substance is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

For the given images:

For a:

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

$K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16$

For b:

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2$

For c:

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

$K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6$

For d:

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}$

Hence, the system which represents the equilibrium having value of $K_c=2.0$ is system (b)

3 0

3 years ago

If you have 222 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?

Dima020 [189]

If the half life is 5 days, you would have to divide 222 by 2 3 times.
so 222/2 =111g
111/2 = 55.5g
55.5/2 = 27.75g

after four half life the mass would be
27.75/2
13.875g

the percent would be
13.875/222
= 6.25%

7 0

3 years ago