Please help due tomorrow Thanks

Iodine-131 has a half-life of 8 days. If the amount of iodine-131 in the orginial sample is 8 g, how much iodine-131 will remain

hram777 [196]

1/32

Explanation it stated with half on 8 days then that means you divide 24 and 8 so its 3 and you have to multiply ✖ ½

7 0

2 years ago

A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.

zysi [14]

Answer:

The initial temperature of the metal is 84.149 °C.

Explanation:

The heat lost by the metal will be equivalent to the heat gain by the water.

- (msΔT)metal = (msΔT)water

-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C

-ΔT = 836.29/12.51 °C

-ΔT = 66.89 °C

-(T final - T initial) = 66.89 °C

T initial = 66.89 °C + T final

T initial = 66.89 °C + 17.3 °C

T initial = 84.149 °C.

7 0

2 years ago

What states of matter were present when pure water reached its boiling point

maxonik [38]

I believe it’s evaporation?

7 0

2 years ago

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?

Sloan [31]

Answer: The concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

Explanation:

Given: pOH = 11.30

The relation between pH and pOH is as follows.

pH + pOH = 14

pH + 11.30 = 14

pH = 14 - 11.30

= 2.7

Also, pH is the negative logarithm of concentration of hydrogen ions.

$pH = - log [H^{+}]$

Substitute the values into above formula as follows.

$pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}$

Thus, we can conclude that the concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

4 0

3 years ago