Match the parts of the nervous system with their functions.

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 2Al2O3 (s) In

jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

6 0

3 years ago

In another experiment, the student titrated 50.0 mL of 0.100 M HC,H,O, with

____ [38]

Answer:

Eqv Pt pH = 8.73

Explanation:

HOAc + NaOH => NaOAc + H₂O

50ml(0.10M HOAc) + 50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O

For neutralized system, 100ml of 0.05M NaOAc remains

NaOAc => Na⁺ + OAc⁻

Na⁺ + H₂O => No Rxn

OAc⁻ + H₂O => HOAc + OH⁻

C(i) 0.05M ----- 0M 0M

ΔC -x ----- +x +x

C(f) 0.05-x

≅ 0.05M ----- x x

Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)

=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M

=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27

pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH

8 0

3 years ago

2. What is the mass of one mole of Na?

borishaifa [10]

Sodium, Atomic mass: 22.989769 g
You can see in a periodic table

5 0

3 years ago

A 1.00-L sample of carbon tetrachloride has a mass of 1.58 kg. What is the density of this substance in g/cm3?

Hoochie [10]

Answer:

1.58gcm⁻³

Explanation:

Given parameters:

Volume of sample = 1.00L

Mass of sample = 1.58kg

unknown:

Density of the sample

Solution:

The density of a substance is expressed as its mass per unit volume.

The units of the parameters are given for volume as liters and for mass as kg.

We need to convert to cm³ for volume and into g for the mass:

1L = 1000cm³:

Therefore, the volume of the sample is 1000cm³

For the mass:

1kg = 1000g

1.58kg gives 1.58 x 1000g = 1580g

The density of the sample will be:

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

Density = $\frac{1580}{1000}$ = 1.58gcm⁻³

3 0

3 years ago

How would you express the frequency of a wave in which five peaks passed a given point each second?

Pani-rosa [81]

n = number of peaks passed a particular point = 5

t = time taken by the peaks to pass a particular point = 1 sec

f = frequency of the wave

by definition , frequency is the repetition of an event in a given interval of time.

we know that frequency is given as

f = n/t

inserting the values

f = 5/1

f = 5 Hz

hence the frequency of the wave comes out to be 5 Hz

3 0

4 years ago