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marin [14]
3 years ago
8

At what temperature does uranium hexafluoride have a density of 0.9560 g/L at 0.5073 atm?

Chemistry
1 answer:
S_A_V [24]3 years ago
8 0
Given:



Density = 0.7360 g/L.

Pressure = 0.5073 atm.
Step 2
The mathematical expression of an ideal gas is,

Chemistry homework question answer, step 2, image 1
Step 3
Here, R is the universal gas constant (0.0821 L-atm/mol K), T is the temperature in Kelvin, and n is the number of
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Determine the pH of a 5x10^-4 M solution of Ca(OH)2
miss Akunina [59]
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-   

Ca(OH)</span>₂ is <span>strong Bases</span><span>

</span>Therefore,  the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]

pOH = - log [ <span>5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!</span>
5 0
3 years ago
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
3 years ago
Compared to the nucleus 5626fe, what is the density of the nucleus 112 48cd?
tia_tia [17]

For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.

n(Cd) / n(Fe)=1

<h3>What is the density of the nucleus 112 48cd?</h3>

Generally, the equation for the density  is mathematically given as

d=\frac{A}{4/3}\piR^3

Therefore

n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3

n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3

n(Cd) / n(Fe)=1

In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same

Read more about density

brainly.com/question/14010194

8 0
2 years ago
Draw the product of the reaction CH3CH2CH2CH2C?CCH3+2Br2??
svetlana [45]
I will assume that the sign ? between the C and the CCH3 is a triple bond, and I will represent it by three vertical lines |||


So the reaction is:


<span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->


This is a typical reaction known as halogenation of alkines.


This is an addition reaction, i.e. the alkyne undergoes an addition of the Br2 (and it also happens with Cl2) to the triple bond to form a tetra halide.


.
                                                                                                      Br   Br
</span>                                                                                                        |     |
<span><span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->  CH3 CH2 CH2 CH2 C - C</span> - CH3
                                                                                                        |     |
                                                                                                        Br   Br


</span>
3 0
3 years ago
4) How many grams are there in 7.40 moles of AgNO3
Ludmilka [50]

Answer:

1260 grams

Explanation:

3 0
3 years ago
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