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mihalych1998 [28]

3 years ago

6

Based on the equation below, how many moles of aluminum sulfate (Al2(SO4)3) will be produced from the reaction of 5.8 moles of a

luminum hydroxide (Al(OH)3) and excess sulfuric acid (H2SO4)?

1 answer:

andrey2020 [161]3 years ago

6 0

Answer:
2.9
Explanation:
1- we can see that for every 2 moles of aluminum hydroxide 1 mole of aluminum sulfate is made so we can make a relation

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taurus [48]

Answer:

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Explanation:

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6 0

2 years ago

Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 8.80 g of lauryl alco

kipiarov [429]

Answer:

The molar mass of lauryl alcohol is approximately 180 g/mol

Explanation:

Step 1: Data given

Mass of lauryl alcohol = 8.80 grams

Mass of benzene = 0.100 kg = 100 grams

Freezing point of Benzene is 5.5 °C

Kf value for benzene = 5.12 °C/molal

Step 2: Calculate the freezing point depression

ΔTf = 5.5 - 3.0 °C = 2.5 °C

Step 3: Calculate molality

⇒ with ΔTf = the freezing point depression = 2.5 °C

⇒ with i = the van't Hoff factor = 1

⇒ with kf = the free point depression constat pf benzene = 5.12°C/m

⇒ with m =the molality = TO BE DETERMINED

molality = ΔTf / kf

molality = 2.5 °C / 5.12 °C /m

molality = 0.488 molal

Step 4: Calculate moles lauryl alcohol

Molality = moles lauryl alcohol / mass benzene

Moles lauryl alcohol = molality * mass benzene

Moles lauryl alcohol = 0.488 m * 0.100 kg

Moles lauryl alcohol = 0.0488 moles

Step 5: Calculate molar mass of lauryl alcohol

Molar mass lauryl alcohol = mass lauryl alcohol/ moles lauryl alcohol

Molar mass lauryl alcohol = 8.80 grams / 0.0488 moles

Molar mass lauryl alcohol = 180 g/mol

The molar mass of lauryl alcohol is approximately 180 g/mol

3 0

3 years ago

1. How many grams are there in 7.5 x 1023 molecules of H2SO4?

Karo-lina-s [1.5K]

Answer:

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5× $10^{23}$ molecules of $H_{2} SO_{4}$ ?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02× $10^{23}$

∴ $\frac{7.5*10^{23} }{6.02*10^{23}}$ =1.25 mol(2d.p.)

Lastly, find the grams using the formula $M= \frac{m}{n}$

m=Mn

m=1.25*98

m=122.5g

-------------------------------------------------------------------------------------------------------------

In conclusion, use those formula to help you:

$M= \frac{m}{n}$ (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: $\frac {molecular mass} {6.02*10^{23}} = moles$

5 0

3 years ago

Which polynomials, given in square inches, could represent the area of a square with whole number side lengths if x is a whole n

erica [24]

<u>Answer:</u> The correct answers are $x^2-9\text{ and }x^2-100$

<u>Explanation:</u>

To find the polynomials which could represent the are of a square having side x greater than 2, we need to find the value of 'x' for all the given polynomials.

From the given options:

1. $x^2-9$

$x^2=9\\x=\sqrt{9}\\x=3,-3$

x = -3 is ignored.

2. $x^2-100$

$x^2=100\\x=\sqrt{100}\\x=\pm10\\x=10,-10$

x = -10 is ignored

3. $x^2-4x+4$

To solve this we use the quadratic formula:

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Putting values of a, b and c, we get:

$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2\times 1}\\x=2,2$

As, x comes out to be 2 and is not greater than 2. Hence, this is not considered.

4. $x^2+10x+25$

Solving for 'x' by splitting the middle term:

$\Rightarrow x^2+5x+5x+25\\\Rightarrow x(x+5)+5(x+5)\\x=-5,-5$

Hence, this is ignored.

5. $x^2+15x+36$

Solving for 'x' by splitting the middle term:

$\Rightarrow x^2+12x+3x+36\\\Rightarrow x(x+12)+3(x+12)\\x=-12,-3$

Hence, this is ignored.

So, the correct polynomials are $x^2-9\text{ and }x^2-100$

3 0

3 years ago

Read 2 more answers

What causes high and low tides?

Klio2033 [76]

I feel like the answer is B, good luck!

3 0

3 years ago