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3 years ago

I need this ASAP! Thank you!

Chemistry

2 answers:

kkurt [141]3 years ago

8 0

Answer:

Explanation:

You look at it straight and at the lowest point the liquid fills to.

stealth61 [152]3 years ago

5 0

35 would be your answer each line is one and it starts at 30 and only goes up 5 so it would be 30

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The boiling point elevation of an aqueous sucrose is found to be 0.39 Celsius. What mass of sucrose (molar mass = 342.30g/mol) w

FromTheMoon [43]

Answer:

130 g of sucrose

Explanation:

Boiling point elevation formula → ΔT = Kb . m

ΔT = Boiling T° solution - Boiling T° pure solvent → 0.39°C

0.39°C = 0.513°C/m . M

m = 0.760 mol/kg → molality = moles of solute / 1kg of solvent

Let's determine the moles of solute → molality . kg

0.760 mol/kg. 0.5 kg = 0.380 moles

If we convert the moles to mass, we'll get the answer

0.380 mol . 342.30 g/mol = 130g

7 0

3 years ago

What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places

lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

−

, not of the hydronium cations,

. In essence, you calculated the

pOH

of the solution, not its

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

mole ratio.

NaOH

(

)

→

(

)

−

(

)

So your solution has

[

−

]

[

NaOH

]

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

−

log

(

[

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

−

log

(

0.150

)

0.824

Now, an aqueous solution at

∘

has

pH + pOH

−−−−−−−−−−−−−−so you can't use

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

−

, not of the hydronium cations,

. In essence, you calculated the

pOH

of the solution, not its

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

mole ratio.

NaOH

(

)

→

(

)

−

(

)

So your solution has

[

−

]

[

NaOH

]

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

−

log

(

[

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

−

log

(

0.150

)

0.824

Now, an aqueous solution at

∘

has

pH + pOH

−−−−−−−−−−−−−−

3 0

2 years ago

According to the law of conservation of energy, the energy released by the system must be transferred to and absorbed by the sur

Andrew [12]

True!
Energy released by system is absorbed by surroundings.

5 0

3 years ago

15.0 mL of an unknown clear liquid is added to a 50 mL graduated cylinder. The mass of the liquid is determined to be 12.7 grams

sveticcg [70]

Answer:

$\boxed {\tt A. \ d=0.85 \ g/mL}$

Explanation:

Density is found by dividing the mass by the volume.

$d=\frac{m}{v}$

The mass of the liquid is 12.7 grams.

We know that 15 mL of this liquid was added to a 50 mL graduated cylinder. Therefore, the volume is 15 mL. The 50 mL is not relevant, it only tells us about the graduated cylinder.

$m= 12.7 \ g\\v= 15 \ mL$

Substitute the values into the formula.

$d=\frac{12.7 \ g}{ 15 mL}$

Divide.

$d=0.846666667 \ g/mL$

Round to the nearest hundredth. The 6 in the tenth place tells us to round the 4 to a 5.

$d \approx 0.85 \ g/mL$

The density of the liquid is about 0.85 grams per milliliter and choice A is correct.

4 0

3 years ago

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

3 years ago