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xenn [34]
3 years ago
9

12. Which element is a metalloid? 1) AI 2) Ar 3) As 4) Au

Chemistry
1 answer:
Nina [5.8K]3 years ago
4 0
I believe it is element As, for arsenic.
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How many liters of radon gas would be in 3.43 moles at standard temperature and pressure (273 K and 100 kPa)?
MissTica

Answer: Option B. 76.83L

Explanation:

1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.

If 1 mole of Radon = 22.4L

Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L

5 0
3 years ago
35.2 J of heat is
ruslelena [56]

The specific heat : c = 0.306 J/g K

<h3>Further explanation</h3>

Given

Heat = 35.2 J

Mass = 16 g

Temperature difference : 7.2 K =

Required

The specific heat

Solution

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Input the value :

c = Q / m.∆T  

c = 35.2 / 16 x 7.2

c = 0.306 J/g K

7 0
3 years ago
Consider the reaction pathway graph below.
Alex17521 [72]
Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!
6 0
4 years ago
Why is conical flask used in the freezing point of naphthalene activity?
SpyIntel [72]

Answer:

D. To ensure the cooling process is not affected by surrounding temperature

Explanation:

The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>

6 0
3 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
3 years ago
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