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Alenkinab [10]
3 years ago
8

Which element is found in laundry products? boron or sulfur

Chemistry
1 answer:
inysia [295]3 years ago
7 0
Boroooooooooooooooon
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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

8 0
3 years ago
If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef
andrezito [222]

Answer:

B

Explanation:

Recall the law of effusion:

\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be <em>r</em>₁ and the rate of hydrogen be <em>r</em>₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for <em>r</em>₂:

\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\  r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}

Because there are 5 moles of hydrogen gas:

\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0
2 years ago
A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0
3 years ago
Which would have the lowest input force?
Afina-wow [57]

Answer:

A long lever with the fulcrum as close as possible to the load

Explanation:

If F be the effort , W be the weight , L₁  be the distance of load from fulcrum and L₂ be the distance of effort from the fulcrum ,

Taking moment of force about the fulcrum , we have

W x L₁ =  F x L₂

F = W x ( L₁ /  L₂ )

F will be minimum when L₁ will be minimum .

Hence  fulcrum should be as close as possible to the load.

5 0
2 years ago
Fossils found in layers of sediment give us clues to Earth's past. Based on the fossil record of this area, we can assume that t
Shalnov [3]

Answer:

its b

Explanation:

i took the test

8 0
3 years ago
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