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3 years ago

10

Causes warm air to rise in the atmosphere. is what type of heat transfer?

1 answer:

shepuryov [24]3 years ago

3 0

Answer:

osmosis is the answer

Explanation:

cause air rises from lower concentration to lower concentration

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. 20 mL ethanol in 60 mL solution

goldenfox [79]

The solution is the mixture of the solute into the solvent particles. In 60 mL solution, 59.8 mL of solvent is added.

<h3>What are solute and solvent?</h3>

Solute and solvent are the components of a solution. They react and mix together to produce a homogenous mixture. The solute is a substance that is added to the solvent.

Solvents are substances that dissolve the added solute in them to make a homogenous solution.

Given,

Volume of Solute = 0.20 mL

Volume of Solution = 60 mL

The volume of solvent is calculated as:

Volume of solvent = Volume of solution - Volume of solute

= 60- 0.20

= 59.8 mL.

Therefore, 59.8 mL of solvent is present in 60 mL of solution.

Learn more about solute and solvent here:

brainly.com/question/20458032

#SPJ1

5 0

2 years ago

Is a Frisbee flying through the air mechanical energy

victus00 [196]

Yes because mechanical energy is the motion of an object. hope that helped!

4 0

3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago

Calculate AGrxn for this equation, rounding your

Agata [3.3K]

Answer: $\Delta G_{rxn}=130.19J$

Explanation:

The balanced chemical reaction is,

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The expression for Gibbs free energy change is,

$\Delta G_{rxn}=\sum [n\times \Delta G_(product)]-\sum [n\times \Delta G_(reactant)]$

$\Delta G_{rxn}=[(n_{CO_2}\times \Delta G_{CO_2})+(n_{CaO}\times \Delta G_{CaO})]-[(n_{CaCO_3}\times \Delta G_{CaCO_3})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta G_{rxn}=[(1\times -394.4)+(1\times -604.17)]-[(1\times -1128.76)]$

$\Delta G_{rxn}=130.19J$

Therefore, the gibbs free energy for this reaction is, +130.19 kJ

4 0

3 years ago

Please help due tonight!!! NO LINKS PLEASE they will be flagged.

Vinvika [58]

Answer:

Carbon fiber if I'm correct

6 0

3 years ago