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3 years ago

Finish the story given the beginning:

Chemistry

1 answer:

Ugo [173]3 years ago

7 0

evaporation will happen so yes the puddles will go away with the sun after awhile.

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What are the missing words?

natka813 [3]

Dont know the first one but
2.particles
3. waves/energy
4.radio/air
5.eardrumbs
6.faster/particles
7.slower/

4 0

3 years ago

Suppose 20.23 g of glucose are dissolved in 95.75 g of water at 27.0 OC. Glucose is nonvolatile (has no vapor pressure) and has

leonid [27]

Answer:

Explanation:

From the information given :

we can understand the solute is glucose and the solvent is water,

So, the weight of glucose = 20.23 g

the molecular weight of glucose = 180.2 g/mol

weight of water = 95. 75 g

the molecular weight of water = 18.02 g/mol

pure vapor pressure of water $P_A = 26.7 \ mmHg$ at 27°C

moles of glucose = weight of glucose/ molecular weight of glucose

= 20.23/180.2

= 0.11 mole

moles of water = weight of water / molecular weight of water

= 95.75/18.02

= 5.31 mole

mole fraction of glucose $X_{glucose} =$ (moles of glucose)/(moles of glucose+ moles of water)

$X_{glucose} =$ 0.11/(0.11 + 5.31)

$X_{glucose} =$ 0.0203

mole fraction of glucose $X_{water} =$ (moles of water)/(moles of water+ moles of glucose)

$X_{water} =$ 5.31/ (5.31 + 0.11)

$X_{water} =$ 0.9797

Using Raoult's Law:

$P_S = P^0_A \times X_A \ \ \ OR \ \ \ P_A = P^0_A \times X_A$

where:

$P_S$ = vapor pressure of the solution

$P_A$ = total vapor pressure of the solution

$P^0_A$ = vapor pressure of the solvent in the pure state

$X_A$ = mole fraction of solvent i.e. water

$P_A =$ 95.75 × 0.9797

$P_A =$ 93.81 mmHg

the total vapor pressure of the solution = 93.81 mmHg

4 0

3 years ago

Can someone help mee pleaseeeee please!

Georgia [21]

Answer:

im sorry dude, just trying to complete my daily challenge

Explanation:

7 0

3 years ago

Write balanced complete ionic equation for hcl(aq)+lioh(aq)→h2o(l)+licl(aq)

olga2289 [7]

H and Li have a +1 charge. Cl and OH have a -1 charge. When written out it should look like this: H(+1) + Cl(-1) + Li(+1) + OH(-1) --> H2O + Li(+1) + Cl(-1) if you wanted the net ionic equation it would be: H(+1) + OH(-1) --> H2O

4 0

4 years ago

Read 2 more answers

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

sweet-ann [11.9K]

Answer:

The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$

Explanation:

The Balanced chemical equation is

$1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)$

Mole ratio of $AgNO_3$ : KCl is 1 : 1

So moles $AgNO_3$ = moles KCl

$Moles KCl = \frac {mass}{molarmass}$

$= \frac {0.785 mg}{(39.1+35.5 g per mol)}$

$= \frac {0.000785 g}{74.6 g per mol}$

$= 0. 0000105 mol KCl$

$= 0.0000105 mol AgNO_3$

So Molarity

$= \frac {moles of solute}{(volume of solution in L)}$

$= \frac {0.0000105 mol}{26.2 mL}$

$=\frac {0.0000105 mol}{0.0262 L}$

= 0.000402M or mol/L is the Answer

(Or) $4.02 \times 10^4 M$ is the Answer

6 0

3 years ago