What are the examples of all the non metals in the first twenty element

Which statement below describes the behavior of a Molecules when a substance changes from a gas to liquid

krek1111 [17]

Answer:

As a substance changes from a solid to a liquid to a gas, its molecules first the molecules are moving fast enough, they are able to "escape." They leave the surface of the liquid as gas molecules. Evaporation is not the only process that can change a substance from a liquid to a gas. The same change can occur through boiling.

Explanation:

hopfully this helps!

7 0

2 years ago

Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2

abruzzese [7]

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

Step 1: Data given

2S (s)+3O2 (g)→2SO3 (g) ΔH = -790.4 kJ

2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ

Step 2: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ

S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g) ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g) ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

4 0

3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Which one of the following will change the value of an equilibrium constant?

Ad libitum [116K]

Answer:

(E) changing temperature

Explanation:

Consider the following reversible balanced reaction:

aA+bB⇋cC+dD

If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:

Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)

where:

[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1

There are some important things to remember when calculating Kc:

- Kc is a constant for a specific reaction at a specific temperature. If you change the reaction temperature, then Kc also changes

- Pure solids and liquids, including solvents, are not considered for equilibrium expression.

- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc

8 0

3 years ago

Describe how you would make 6.0 L of a 0.100 M KCl solution from a 4.0 M stock KCl solution.

Galina-37 [17]

Answer:

0.15 l of 4.0 m stock KCl solution should betaken

Explanation:

N1V1=N2V2

6*0.1=V2*4

V2=0.15L

5 0

2 years ago

What are the examples of all the non metals in the first twenty element​

What are the examples of all the non metals in the first twenty element