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Masteriza [31]
2 years ago
13

If two unidentified solids of the same texture and color have different solubilities in 100 grams of water at 20°C, you could co

nclude that
a.
they are the same substance.
b.
they are different substances.
c.
they have different melting points.
d.
their solubilities will be the same if the water temperature is increased.
Chemistry
2 answers:
AlekseyPX2 years ago
3 0
I'm certain it's "D"

...because it can't be "A" or "B" because solubility IS a property but to actually determine whether these two substances are the same or different we would need at least two-three properties (like boiling point or specific heat).

and it can't be "C" because the melting point is just simply irrelevant when comparing the solubility of two substances.
Licemer1 [7]2 years ago
3 0

they have different melting points.

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2 years ago
Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm
ivolga24 [154]

Answer:

O_{2} and NO_{2}

Explanation:

For a given system at constant temperature, the number of moles of gas present in the system is proportional to the product of the system pressure and volume. Therefore, we have:

NO: 6 L * 0.7 atm = 4.2 L*atm

O:  1.5 L* 2.5 atm = 3.75 L*atm

For the given system based on a balanced chemical equation:

2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e. P_{NO} = 0

For the product: 2.70 L*atm NO produced  2.70 L*atm NO_{2}

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of NO_{2}  = (2.70 L*atm NO_{2} ) / (7.5 L) = 0.36 atm NO_{2}  

Similarly for oxygen gas:

3.75 L*atm - 1.35 L*atm  = 2.40 L*atm oxygen gas remaining  

Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm  

Thus, the gases present at the end of the experiment are O_{2} and NO_{2}

3 0
2 years ago
A solution of 0.108 M cysteine is titrated with 0.0540 M HNO3 . The pKa values for cysteine are 1.70 , 8.36 , and 10.74 , corres
kirill [66]

Answer:

2.698

Explanation:

Cysteine + H+ <==> cysteineH+

at the equivalence point

cosidering 1 L of cysteine solution

volume of HNO3 to reach equivalence point = 0.108 M x 1 L/0.0540 M = 2.0 L

[CysteineH+] formed = (0.108 M x 1 L)/(1.0 + 2.0) L = 0.036 M

hydrolysis of salt

cysteineH+ + H_2O <==> cysteine + H_3O+

let x amount hydrolyzed

Ka1 = [cysteine][H3O+]/[cysteineH+]

pKa1 = -log[Ka1] = 1.70

Ka = 0.02

so,

0.02 = x^2/(0.036-x)

x^2 + 0.02x - 0.00072 = 0

x = 0.002

pH at equivalence point = -log(0.002) = 2.698

5 0
2 years ago
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