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Elis [28]
3 years ago
10

In the solubility lab, what did you change the temperature of

Chemistry
1 answer:
NISA [10]3 years ago
6 0
I'm taking the test right now; I'll edit this when I find out if I was right! (I chose "solvent" but I wanna be sure.)

Edit: Solvent is the correct answer!
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What is flammability?
Masteriza [31]
Well, Flammability is: What happens when combustible liquids (liquids that can burn) ignite (catch on fire) and burn easily at normal working temperatures.

Hope I helped!

- Amber
3 0
3 years ago
Read 2 more answers
What is the value of the equilibrium constant for this redox reaction?
aivan3 [116]
Correct  answer: Option D, <span>K = 5.04 × 10^52
</span>
Reason:
We know that, 
Ecell = \frac{0.0592}{n}log(K),
where n = number of electrons = 2 (in present case)
K = equilibrium constant.

Also, Ecell = <span>+1.56 v

Therefore, 1.56 = </span>\frac{0.0592}{2}log(K)
Therefore, log (K) = 52.703
Therefore,  K = 5.04 X 10^52


8 0
3 years ago
Read 2 more answers
The normal freezing point of a certain liquid
slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g  = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X = 0.4^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

K_f = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}

K_f=4.12^oC/m

Therefore, the molal freezing point depression constant of liquid X is, 4.12^oC/m

3 0
3 years ago
The reaction na 3po 4( aq) 3 agno3( aq) → ag 3po 4( s) 3 nano 3( aq) is best classified as a(n):_____.
Ugo [173]

The reaction  Na_{3} PO_{4} ( aq) +3AgNO_{3} (aq) → Ag_{3} PO_{4} ( aq) +3NaNO_{3} (aq) is best classified as double displacement reaction.

Those reaction in which two compounds react by exchanges of ions to form two new compounds is called  double displacement reaction. The easiest way to identify double displacement reactions is to check to see whether the cations exchanged anions with each other  or not . Always balanced chemical equation is used to determine.

There are three types of double displacement reaction which is given as,

  1. Precipitation
  2. Neutralization
  3. Gas formation

The real world example of double displacement reaction is combining vinegar and baking soda to create homemade volcano.

learn more about double displacement reaction

brainly.com/question/13870042?

#SPJ4

6 0
1 year ago
Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil
quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

7 0
3 years ago
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