Is air matter or energy

Which of the following molecules are considered inorganic?

Oksi-84 [34.3K]

The answer is D water

7 0

2 years ago

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio

zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate $SO_4~^-^2$ and the chloride $Cl^-$ :

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.( $Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2$ ), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ( $Pb^+^2~,Ag^+~,Hg_2~^+^2$ ), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

7 0

3 years ago

A block of aluminum occupies a volume of 15.0 mL and weighs 45 grams.

vladimir2022 [97]

Answer: density equals 3 g/mL

Step by step explanation:

D=m/v
D=45/15
D=3

5 0

3 years ago

Which of the following statements provides the BEST explanation for the many chemical similarities between two elements in a Mai

rjkz [21]

Answer:

(b) Both have the same number of valence electrons.

Step-by-step explanation:

We find the most striking chemical similarities between two Main Group elements when they are in the same Group of the Periodic Table.

Elements in the same Group have the same number of valence electrons.

(a) is wrong, because elements in the same group have different masses.

(c) is wrong, because atoms with the same number of protons belong to the same element.

(d) is wrong, because elements in the same Group must be in . different Periods.

7 0

3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

Learn more:

5 0

2 years ago