Ask question

Ask question

All categories

2 years ago

9

Which of the following is NOT a limiting factor for the number of possible compounds?

1 answer:

babymother [125]2 years ago

6 0

Answer:

all elements have a different atomic number

Explanation:

Atomic number of element does not affect their reactions with others

You might be interested in

A 1.000-g sample of lead shot reacted with oxygen to give 1.077 g of product. Calculate the empirical formula of the lead oxide

Evgesh-ka [11]

It mean it consisted of 1 g of lead and 0.077 g of O2.
divide these numbers by molar mas.
1/82=0.012 Pb /0.004 = 3
0.077/16= 0.004 O /0.004 =1
Pb3O

3 0

2 years ago

Read 2 more answers

A compound decomposes by a first-order process. if 25.0% of the compound decomposes in 60.0 minutes, the half-life of the compou

Tom [10]

Answer:<span>d. 145 minutes
</span>
Half-life is the time needed for a radioactive to decay half of its weight. The formula to find the half-life would be:

Nt= N0 (1/2)^ t/h

Nt= the final mass
N0= the initial mass
t= time passed
h= half-life

If 25.0% of the compound decomposes that means the final mass would be 75% of initial mass. Then the half-live for the compound would be:
Nt= N0 (1/2)^ t/h
75%= 100% * (1/2)^ (60min/h)
3/4= 1/2^(60min/h)
log2 3/4 = log2 1/2^(60min/h)
0.41503749928 = -60min/h
h= -60 min / 0.41503749928= 144.6min

3 0

3 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][Br^-]$

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][Br^-]$

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %

3 0

2 years ago

How many moles are in 20 grams of Ca (Calcium)? *

Flura [38]

Answer:

0.49903 mol

Explanation:

5 0

2 years ago

What us a bond between a positive and a negative ion called

Nat2105 [25]

Static. Is the bond if u need 2 know

6 0

2 years ago