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telo118 [61]
3 years ago
13

When two your substances are combined so that each of the pure substances retains its own properties the result is a(n)?A. mixtu

reB. isotopeC. elementD. compound
Chemistry
1 answer:
goblinko [34]3 years ago
8 0
Mixture is the answer 
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I need the answers? Pls help!
Vesnalui [34]
Im pretty sure its h2o so like the 2nd one
4 0
2 years ago
The phrase “the danger is in the dose” is sometimes used when describing poisonous (toxic) chemicals. What does this mean?
Lostsunrise [7]
It means it depends on how much of the chemical you have. For ex. if 1 drop of snake venom might not kill you, but 20 drop could definitely kill you. Hope this helps!
6 0
3 years ago
What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
mart [117]

Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

Molar mass of NaCl = 58.44 g/mol

Mass = Moles*Molar mass = 0.0075\times 58.44\ g = 0.4383 g

6 0
3 years ago
Assuming 100% dissociation, calculate the freezing point and boiling point of 1.22 m SnCl₄(aq). Constants may be found here.
Gekata [30.6K]

Answer:

T° freezing solution → -11.3°C

T° boiling solution → 103.1 °C

Explanation:

Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"

This salt dissociates as this:

SnCl₄ (aq)  →   1Sn⁴⁺ (aq)  +   4Cl⁻  (aq)   (so i =5)

The formula for the colligative property of freezing point depression and boiling point elevation are:

ΔT = Kf . m . i

where ΔT = T° freezing pure solvent - T° freezing solution

ΔT = Kb . m . i

where ΔT = T° boiling solution - T° boiling pure solvent

Freezing point depression:

0° - T° freezing solution = 1.86°C/m . 1.22 m . 5

T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C

Boiling point elevation:

T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5

T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C

8 0
3 years ago
An acid has an acid dissociation constant of 2.8x10^-9. What is the base dissociation constant of its conjugate base?
jok3333 [9.3K]
PH=-lg[H⁺]
pH=-lg(2.8×10⁻⁹)=8.553

pOH=14-pH
pOH=14-8.553=5.447

[OH⁻]=10^(-pOH)
[OH⁻]=10⁻⁵·⁴⁴⁷=3.6×10⁻⁶
3 0
3 years ago
Read 2 more answers
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