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3 years ago

Here’s another one guys please help.

Chemistry

1 answer:

ch4aika [34]3 years ago

3 0

Answer:

Explanation:

i think B im not sure

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Turn on Write equation. What you see is an equation that shows the original uranium atom on the left. The boxes on the right rep

My name is Ann [436]

Answer:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

$_{92} ^{238} U_{}$ → $_{90} ^{234} Th_{} + _{2} ^{4} He_{}$

I hope this explanation is clear and explanatory.

6 0

3 years ago

What is conjugate solution

mars1129 [50]

Answer:

Explanation: A mixture of two partially miscible liquids

5 0

3 years ago

2. What pressure is required to compress 196.0 L of air at 1.00 atmosphere into a cylinder

damaskus [11]

Answer:

Explanation:

The required pressure can be found by using the formula for Boyle's law which is

$P_2 = \frac{P_1V_1}{V_2} \\$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

From the question we have

$P_2 = \frac{1 \times 196}{26} = \frac{196}{26} \\ = 7.538461...$

We have the final answer as

Hope this helps you

3 0

2 years ago

At the beginning of an experiment, a scientist has 176 grams of radioactive goo. After 165 minutes, her sample has decayed to 5.

Paul [167]

The half-life equation is written as:

An = Aoe^-kt

We use this equation for the solution. We do as follows:

5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE

Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?

G(t) = 176e^-0.02t

How many grams of goo will remain after 50 minutes?

G(t) = 176e^-0.02(50) = 64.75 g

6 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago