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deff fn [24]
3 years ago
10

A 2.00 g sample of a bromine oxide (brxoy) is converted to 2.94 g of agbr. all the bromine in the original oxide compound ends u

p in the agbr (molar mass for agbr = 187.8). determine the empirical formula of the bromine oxide.
Chemistry
1 answer:
qwelly [4]3 years ago
4 0
In comparing two isotopes of the same element, the atomic number, stays the same.

Allow me to explain: The protons in an isotope ALWAYS stays the same, but the neutrons change. But, what's the atomic number made up of? 

Protons, which means it won't change. 

However, the atomic mass is made up of neutrons + protons, so only the atomic mass would change - not the atomic number.  
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What is the pH of a solution with a concentration of 4.7 × 10-4 molar H3O+?
wlad13 [49]
The pH of a liquid substance is calculated through the equation,
                                       pH = -log[H3O+]
Substituting the given concentration of the hydronium ion to the equation above,
                                        pH = -log[4.7x10^-4 M]
The value of pH is equal to 3.33. Thus, the pH of the solution is approximately 3.33. 
3 0
3 years ago
What is the pH of this solution?
Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M

Which is also:

[OH^-]=1.643x10^{-3}M

Thereafter we can compute the pOH first:

pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784

Finally, the pH turns out:

pH=14-2.784\\\\pH=11.216

Regards!

5 0
3 years ago
what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?
nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0
2 years ago
The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri
Nimfa-mama [501]

Answer: The expression for equilibrium constant is \frac{[NH_3]^2}{[H_2]^3[N_2]}

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as k_c

For a general reaction:

aA+bB\rightleftharpoons cC+dD

The equilibrium constant is written as:

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

Chemical reaction for the formation of ammonia is:

N_2+3H_3\rightleftharpoons 2NH_3

k_c=1.6\times 10^2

Expression for k_c is:

k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}

1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}

8 0
3 years ago
In a nuclear reactor, when the control rods are lowered between the fuel rods, they slow the fission reactions.
rosijanka [135]
True when the boron control rods are lowered it slows the reaction
5 0
3 years ago
Read 2 more answers
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