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3 years ago

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Chemistry

1 answer:

zmey [24]3 years ago

8 0

Answer:

d. 12.3 grams of Al2O3

Explanation:

Based on the reaction:

4Al + 3O2 → 2Al2O3

<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>

<em />

To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:

<em>Moles Al -Molar mass: 26.9815g/mol-</em>

6.50g * (1mol / 26.9815g) = 0.241 moles Al

<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>

0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3

0.120 moles Al2O3 * (101.96g / mol) =

12.3g of Al2O3 are produced.

Right answer is:

<h3>d. 12.3 grams of Al2O3 </h3>

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We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a

sukhopar [10]

Answer:

See Explanation

Explanation:

$\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n$

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

= 236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) $\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy$

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0

2 years ago

In a titration experiment 12.5 ml of 0.500 m h2so4 neutralized 50.0 ml of naoh. the concentration of the naoh solution is ____.

s344n2d4d5 [400]

0.250 mol/L

<em>Step 1</em>. Write the chemical equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

<em>Step 2</em>. Calculate the moles of H2SO4

Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)

= 6.25 mmol H2SO4

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)

= 12.5 mmol NaOH

<em>Step 4</em>. Calculate the concentration of the NaOH

[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L

4 0

3 years ago

What is the molarity if 6664 grams of boron iodide is dissolved in 5.25 liters of water?

Virty [35]

Answer:

3.24 mol/L

Explanation:

Given that:

mass of Boron triiodide = 6664 grams

molar mass of BI_3 = 391.52 g/mol

Recall that:

number of moles = mass/molar mass

∴

number of moles = 6664 g /391.52 g/mol

number of moles = 17.02 mol

Also;

Molarity = moles for solute/liter for solution

= 17.02 mol/5.25 L

= 3.24 mol/L

4 0

2 years ago

If i combined 15.0 grams of calcium hydroxide with 75.0 ml of 0.500 m hcl, how many grams of calcium chloride would be formed?

Zina [86]

The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>

n=mass in g/M.M

15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂

no. of mol of HCl:

n=0.5 mol/L*0.075L=0.0375 mol

This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.

Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.

Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂

no. of mol of CaCl₂= 0.0375/2= 0.01875 mol

mass in g=n*MM= 0.01875*111= 2.08 g

7 0

3 years ago

A piece if stone has a mass of 24.595 grams and a volume of 5.34 cm ^3. What is the density of the stone

Rudik [331]

Answer:

The answer is

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of stone = 24.595 g

volume = 5.34 cm³

The density of the stone is

$density = \frac{24.595}{5.34} \\ = 4.605805243...$

We have the final answer as

Hope this helps you

5 0

2 years ago