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3 years ago

What is true about two neutral atoms of the element gold

1 answer:

3 0

I'm pretty sure it's they have the same properties, because they don't have different masses, they don't both have missing nuclei, and they don't hav a different # of protons. Can't recall much else about the two, but I distinctly remember that

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A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+

natima [27]

Answer:

A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .

(4x^2)x

Kc= -----------

(A-2x)^2

PV=nRT

n/v = P/RT = .745/(0.0821)(334+273) = .01495

To Find the initial molarity of NO2

(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L

Thus:

46(A-2x) + 2x(30) + 32x = .515 g/L

46A-92x+60x+32x = .515

46A=.515

A=.01120 M

Using the total molarity found

(A-2x)+2x+x = .01495 M

A+x=.01495

Plug in A found into the above equation:

.01120+x = .01495

x=.00375

Now Plug A and x into the original Equilibrium Constant Expression:

(4x^2)x

Kc= -----------

(A-2x)^2

Kc = 0.000014

Explanation:

3 0

3 years ago

A spoonful of salt is stirred into a pot of water.The salt is a

pshichka [43]

Answer:

the salt is a solute in this instance

4 0

3 years ago

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago

What is the percent by mass of magnesium sulfate in MgSO4 · 7H2O?

ad-work [718]

The breakdown of mass percentage of the elements is determined by dividing molar masses to the total molar mass of the compound (246.37 g/mol). There are four elements: for Mg: 9.87%, S:13.01%, H:5.69%, 71.44%.

5 0

3 years ago

During the phosphatase experiment you will use a 1% w/v of phenolphthalein di-phosphate (PPP). How much PPP do you need to make

MrMuchimi

Answer:

To make 500 mL of a solution of PPP 1% w/v you need 5g

Explanation:

A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w):

% w/v = g of solute/ mL of solution × 100

A 1% w/v means that you have 1 g of PPP per 100 mL of solution

Thus, if you need to make 500 mL of solution you will need 5 g of PPP.

I hope it helps!

7 0

3 years ago