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Svetradugi [14.3K]
3 years ago
10

A chemist dissolved crystals of an unknown substance into water at room temperature. He found that 33 g of the substance can be

dissolved into 100 mL of water. What property of the unknown substance was the chemist most likely investigating? its solubility its rate of dissolution its chemical reaction to water its reaction to temperature changes
Chemistry
2 answers:
Inga [223]3 years ago
8 0
The answer is: It's solubility
Anon25 [30]3 years ago
8 0
He was testing for solubility
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How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

6 0
2 years ago
An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of
seraphim [82]

Answer:

The temperature difference of the body after 3 hours = 5.16 K

Explanation:

we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹

                                   or, 1.2 mole.h⁻¹

The average heat evolved by the oxidation of foodstuffs is then:

⇒          Q avg =\frac{1.2 X 420 X 10^{3} }{70} = 7.2 kj.h⁻¹.Kg⁻¹

the heat produced after 3 h would be:

                 =    7.2 kj. h⁻¹.Kg⁻¹ x 3 h

                 = 21.6 kj. kg⁻¹

                 = 21.6 x 10³ j kg⁻¹

We know Qp = Cp x ΔT

Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹

⇒ ΔT = \frac{Qp}{Cp}

⇒ ΔT = \frac{(21.6 X 10^{3} j.kg^{-1} ) }{(4.18 j k^{-1}g^{-1})   X (1000g.kg^{-1} )}

⇒ ΔT = 5.16 K

6 0
3 years ago
Consider these elements: p, ca, si, s, ga.<br> a. write the electron configuration for each element
dimulka [17.4K]
Method:

1) Find the atomic number in a periodic table: the number of electrons equal the atomic number

2) Use Aufbau rule

Element     atomic number       electron configuration
<span>
P                15                            1s2 2s2 2p6 3s2 3p3

Ca              20                            </span><span><span>1s2 2s2 2p6 3s2 3p6 4s2

</span>Si                14</span><span>                            1s2 2s2 2p6 3s2 3p2

S                 16</span><span><span>                            1s2 2s2 2p6 3s2 3p4

</span>Ga               31.                   </span><span><span>        1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p</span>       </span>
7 0
3 years ago
Please help with 14-16 <br> Thank you!
Natalka [10]
I don't know about 14, but 15 is (4), because a liquid draws in heat to turn into a gas. 16 is (2), because to turn into a cold solid, something has to release heat.
4 0
3 years ago
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