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3 years ago

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A) A cat dish has a mass of 1975 g and a volume of 7500ml. What is the overall density of the cat dish? B) find the maximum mass

that the cat dish can hold before it sinks

1 answer:

maxonik [38]3 years ago

5 0

What's the answer? It asked to be 20 characters long so just writing this.

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.

wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>

5 0

3 years ago

Chloroform is a liquid once used for anesthetic. What is the volume of 5.0 g of chloroform? The density of chloroform 1.49 g/mL

mario62 [17]

Answer:

3.3557047 mL

Explanation:

The density can be found using the following formula:

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

Let's rearrange the formula to find the volume, $v$ .

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

$d*v=m$

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

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

The volume can be found by dividing the mass by the density. The mass of the chloroform is 5 grams and the density is 1.49 grams per milliliter. Therefore,

$m= 5g \\d= 1.49 g/mL$

Substitute the values into the formula.

$v=\frac{5g}{1.49 g/mL}$

Divide. When we divide, the grams, or g, in the numerator and denominator will cancel out.

$v= \frac{5}{1.49}mL$

$v=3.3557047 mL$

The volume of 5 grams of chloroform is 3.3557047 milliliters

4 0

3 years ago

A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa

Phoenix [80]

Answer:

$V_{2} = 8.92 L$

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

$P_{1} V_{1} =nRT$

$320Kpa*0.003m^{3} =nRT$

$1000L = 1m^{3}$

So

$nRT= 0.96$

Now, as for the final condition:

$P_{2}V_{2}=nRT$

$P_{2} V_{2} =0.96$

clearing $V_{2}$

$V_{2} =\frac{0.96}{P_{2} }$

$V_{2} =0.00872m_{3}$

$V_{2} = 8.92 L$

6 0

3 years ago

What is the hydrogen-ion concentration of the ph is 3.7

Komok [63]

Answer:

the answer 37

Explanation:

4 0

3 years ago

When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed? (2 points) 2Al

barxatty [35]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

5*3 = 15 moles of HCl needed to fully react with 5 moles of Al. Since 13<15, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced so choice B.

Hope this helps!!! :)

5 0

4 years ago