1. In addition to being necessary for respiration, _______ also provides UV radiation protection.

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

Explanation :

The balanced chemical reaction will be:

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

First we have to calculate the moles of Al.

$\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}$

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{55g}{27g/mol}=2mole$

Now we have to calculate the moles of $O_2$ .

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of $O_2$

So, 2 mole of Al react with $\frac{3}{4}\times 2=0.75\times 2$ moles of $O_2$

Now we have to calculate the volume of $O_2$ consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of $O_2$ occupies 22.4 liter volume of $O_2$ gas

So, $0.75\times 2$ mole of $O_2$ occupies $0.75\times 2\times 22.4$ liter volume of $O_2$ gas

Therefore, the volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

3 0

3 years ago

Water is added to liquid isopropanol (a polar liquid) to form a solution of rubbing alcohol. Describe what happens on the molecu

olya-2409 [2.1K]

Explanation:

Since liquid isopropanol is a polar liquid and water is also a polar solvent. So, when both of them are added together then according to the like dissolves like principle they get dissolved.

At the molecular level, the polar molecules of isopropanol get attracted towards the polar molecules of water at the surface of water.

As a result, water molecules get surrounded by isopropanol. Thus, water molecules enter the solution and evenly spread into the solution.

5 0

3 years ago

Read 2 more answers

Help would be majorly appreciated

Sophie [7]

Answer:

a) the atomic number is 15

b) the mass number is 15+16 = 31

c) element is phosphorus

d)Group 15 period 3

6 0

3 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

3 years ago