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3 years ago

5

Which of these is an example of a physical change?

2 answers:

oee [108]3 years ago

7 0

I'm pretty sure its liquid water turning into vapor.

igomit [66]3 years ago

4 0

Liquid water turning into vapor.

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The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago

Knowing that all three-sided figures are triangles is an example of:

pogonyaev

Answer:

B. Concept Formation

Explanation:

All the other answer choices are not equivalent.

Hope this helps!

Brainliest??

5 0

3 years ago

Read 2 more answers

The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga

Rashid [163]

Answer:

Final temperature of the gas is 576 $^{0}\textrm{C}$ .

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

where $V_{1}$ and $V_{2}$ are volume of gas at $T_{1}$ and $T_{2}$ temperature (in kelvin scale) respectively.

Here $V_{1}=158mL$ , $T_{1}=(273+25)K=298K$ and $V_{2}=450mL$

So $T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K$

849 K = (849-273) $^{0}\textrm{C}$ = 576 $^{0}\textrm{C}$

So final temperature of the gas is 576 $^{0}\textrm{C}$ .

3 0

3 years ago

Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.

Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= $moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.241}{0.0137kg}=17.6mole/kg$

Therefore, the molality of solution is 17.6 mole/kg

3 0

3 years ago

A container with 3.0 moles of gas has a volume of 60.0L with a temperature at 400.K what is the pressure

iris [78.8K]

Answer: P= 1.64 atm

Explanation: solution attached.

Use Ideal gas law

PV= nRT

Derive for P

P= nRT/V R= 0.08205 L.atm/mol.K

Substitute the values.

3 0

3 years ago