A volcano ejects a glowing chunk of lava straight up.

If atoms of a halogen nonmetal (Group 17) gains one electron, the atoms the have __.

RUDIKE [14]

Hm, this could be more than one option, but gaining electrons makes a negative charge, so

If atoms of a halogen nonmetal (Group 17) gains one electron, the atoms the have "a negative one charge".

8 0

3 years ago

Read 2 more answers

What is A collection of waves called

sattari [20]

When waves act together, you talk about "interference". When they reinforce each other, it is "constructive interference". When they cancel each other, it is "destructive interference".

5 0

4 years ago

How many moles of silver can be produced from silver nitrate from 1 mole of zinc?

jasenka [17]

Answer:

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

Zn + 2AgNO3 ---> Zn (NO3)2 +2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = 1/2 x 6.5 = 3.25 moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = 2/2 x 6.5 = 6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

4 0

2 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago

Show work, thanks

GarryVolchara [31]

Answer:

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution

7 0

3 years ago