How many moles are in 75.0g of Na2CO3

What net force is required to cause a 12,000 kg airplane to accelerate at a rate of 5.2 m/s

galina1969 [7]

The answer is 624000N

6 0

2 years ago

The outputs of celluar respritiation are?

-BARSIC- [3]

Answer:

carbon dioxide , water , 38ATP

Explanation:

There are two types of respiration:

1. Aerobic respiration

2. Anaerobic respiration

Aerobic respiration

It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.

Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.

Glucose → lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.

5 0

3 years ago

Three Stoichiometry Questions

andrezito [222]

Answer:

Explanation:

7)

Given data:

Mass of aluminium = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with aluminium and oxygen.

Al ; Al₂O₃

4 : 2

0.09 : 2/4×0.09 = 0.045

O₂ : Al₂O₃

3 : 2

0.08 : 2/3 ×0.08 = 0.053

The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.

Mass of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.045 × 101.96 g/mol

Mass = 4.6 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.5 g / 4.6 ×100

Percent yield = 76.1%

8)

Given data:

Mass of copper produced = 3.47 g

Mass of aluminium = 1.87 g

Percent yield = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 1.87 g/ 27 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of copper with aluminium.

Al ; Cu

2 : 3

0.07 : 3/2×0.09 = 0.105

Mass of copper:

Mass = number of moles × molar mass

Mass = 0.105 × 63.55 g/mol

Mass = 6.67 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.47 g / 6.67 × 100

Percent yield = 52%

4 0

3 years ago

What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)

astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH, $10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})$

= 0.002 mol

For Cinnamic acid, $30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})$

= 0.003 mol

Acid and base react as:

$HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)$

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$