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damaskus [11]
3 years ago
12

How to find the solute and solvent

Chemistry
2 answers:
Galina-37 [17]3 years ago
4 0
The person above is 100% correct. I would write it but that would be me copying her/his answer
Salsk061 [2.6K]3 years ago
3 0

the solute is the one that dissolves meaning its particles are separating into the solvent, and the solvent is the one that dissolves the other substance.

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A balloon at 25°C has 30 L. What will the balloon's volume at 35°C?
anygoal [31]

The balloon's volume at 35°C : V₂=31.01 L

<h3>Further explanation</h3>

Given

T₁ = 25°C+273 = 298 K

V₁ = 30 L

T₂ = 35 °C + 273 = 308 K

Required

The new volume (V₂)

Solution

Charles's Law  

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Input the value :

V₂=(V₁.T₂)/T₁

V₂=(30 x 308)/298

V₂=31.01 L

8 0
3 years ago
Which tool is used to measure mass?
slega [8]

Answer:

kilogram is used to measure to mass

3 0
3 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
Which of the following statements is true with regard to transverse and longitudinal waves?
Ipatiy [6.2K]

Answer:

D (The last answer)

Explanation:

In a transverse wave, particles oscillate perpendicular to the direction of wave motion.

In a longitudinal wave, the oscillations of particles are parallel to the direction of propagation.

5 0
4 years ago
Read 2 more answers
Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium
Elodia [21]

Answer: sodium amide undergoes an acid -base reaction

Explanation:

sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as  there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

This leads to formation of ammonia and sodium methoxide.

Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

4 0
3 years ago
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