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allochka39001 [22]
3 years ago
13

Why do the aquatic plants not break down by water current?

Chemistry
1 answer:
Lelechka [254]3 years ago
6 0
Leaves of aquatic plants are covered by a waxy cuticle which does not allow the water to wet them.
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How many sig figs are in 100.00?
d1i1m1o1n [39]

Answer:

5

Explanation:

6 0
3 years ago
Need help !!!!! ASAP
Julli [10]
<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

V_{2}=1L

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

P_{1}V_{1}=P_{2}V_{2}

So, we are given the information:

V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa

Then, isolating the new volume and substituting into the equation, we have:

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\frac{P_{1}V_{1}}{P_{2}}

V_{2}=\frac{50kPa*2L}{100kPa}=1L

Hence, the new volume will be 1 L.

V_{2}=1L

Have a nice day!

4 0
3 years ago
Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -
scoray [572]

Answer:

Part 1: - 1.091 x 10⁴ J/mol.

Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

6 0
4 years ago
Read 2 more answers
Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. ????????????H(????) → ????????+(?????
Dennis_Churaev [7]
Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)

Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
5 0
3 years ago
1) 120KM/H A M/S Y CM/S
adell [148]
4) would be your correct answer
8 0
3 years ago
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