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2 years ago

Select each process that shows a physical change. Select all that apply.

Chemistry

2 answers:

erica [24]2 years ago

5 0

Answer:

Chopping wood logs

A pot of water o a grate over a burning fire

Explanation:

Physical change is easily reversible. Burning/combustion is a chemical process where substances react rapidly with oxygen: this is usually irreversible.

The marshmallow, roasted food and burned wood all undergo combustion and hence are tagged chemical changes.

Explanation:

lakkis [162]2 years ago

4 0

Answer:

chopping wood is the answer

Explanation:

burning, heating , roasting are all chemical changes because the chemical properties are changed. hope this helps!!!!

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Often express in psi, pressure is the term for force per unit of

taurus [48]

Answer:

Surface area

Explanation:

psi = pound square inch

Often expres in psi, pressure s the term for force per unit of surface area.

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1 year ago

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Agata [3.3K]

Answer:

Osmosis is the diffusion of ⇒ water across a selectively permeable membrane. This process does not require the cell to use ⇒ energy to move molecules. It is an example of ⇒ passive transport.

Explanation:

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3 years ago

Read 2 more answers

Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What i

AlekseyPX

Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

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3 years ago

Can someone please answer questions 32 and 33?

Talja [164]

32.) 55.6 mL HCl
33.) 0. 128 M

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3 years ago

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

photoshop1234 [79]

Answer:

Approximately $8.38 \times 10^8\; \rm mL$ , which is the same as $8.38 \times 10^5 \; \rm L$ . Assumption: the behavior of this gas is ideal.

Explanation:

Initial state of the gas:

$T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K$ .
$P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^{12}\; \rm Pa$ .

STP state:

$T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K$ .
$P_\text{STP} = 10^5\; \rm Pa$ .

The initial state of this gas can be changed to STP state in two steps:

First, reduce the pressure from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ .
Second, increase the temperature from $260.65\; \rm K$ to $273.15\; \rm K$ .

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ . By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL \times \frac{10^{12}\; \rm Pa}{10^5\; \rm Pa} = 8.00 \times 10^8\; \rm mL\end{aligned}$ .

After that, assume that pressure stays the same when the temperature changes from $\rm 260.65\; \rm K$ to $\rm 273.15\; \rm K$ . By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 \times 10^8\; \rm mL \times \frac{273.15\; \rm K}{260.65\; \rm K} \\ &\approx 8.38\times 10^8\; \rm mL = 8.38 \times 10^5\; \rm L\end{aligned}$ .

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3 years ago