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makkiz [27]
3 years ago
9

When chlorine gas is passed through a potassium bromide solution, bromine forms in a potassium chloride solution?

Chemistry
1 answer:
pshichka [43]3 years ago
4 0
Yes because chlorine is more reactive than bromine, it displaces bromine in potassium bromide solution.
Forms bromide *gas* in KCl solution
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When the pressure that a gas exerts on a sealed container changes from 22.5 psi to ? psi, the temperature changes from 110 degre
natta225 [31]
Using Gay-Lussac's Law, pressure is proportional to (absolute) temperature in Kelvin. We first convert the temperature values to Kelvin: 110 C = 383.15 K, while 65 C = 338.15 K.
P1/T1 = P2/T2
22.5/383.15 = P2/338.15
P2 = 19.9 psi
8 0
3 years ago
A block is found to have volume of 35.3 cm3. It’s mass is 31.7g. Calculate the density of the block
11111nata11111 [884]

Answer:

<h2>1.11 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{35.3}{31.7}  \\  = 1.113564

We have the final answer as

<h3>1.11 g/mL</h3>

Hope this helps you

6 0
2 years ago
Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
An acid is used to bring the pH of a solution from pH 9 to pH 6. How many times more acidic is the final solution than the initi
Molodets [167]
Bleh bleh bleh bleh bleh bleh bleh bleh and bleh
5 0
2 years ago
The charge on an ion is known as its___number.
a_sh-v [17]

Answer:

oxidation number is correct!! :)

Explanation:

8 0
2 years ago
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