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2 years ago

Give an example of a base in chemistry

Chemistry

2 answers:

slega [8]2 years ago

8 0

An example of a base in chemistry would be like soap or lye

vivado [14]2 years ago

4 0

Acetone which is a weak Lewis base. Acetone: C3H6O

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Hydrates are formed with _______ compounds

Triss [41]

Answer:

ionic compounds

Explanation:

7 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago

en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es

Alinara [238K]

The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

<em />

solution 1

composition :

35% acid :

$\tt 0.35\times 800~g=280~g$

water :

$\tt 800-280=520~g$

solution 2(new solution)

composition :

water

$\tt 520-(80~ml\times 1~g/ml)=440~g$

Total mass of new solution after water evaporated

$\tt 280(acid)+440(water)=720~g$

%mass of acid in a new solution

$\tt \dfrac{280}{720}\times 100\%=38.9\%$

5 0

2 years ago

2C2H2 + 5O2 → 4CO2 + 2 H2O How many grams of CO2 are required to react with exactly 3.00 mol of O2? 106 g 132 g 165 g 76.8 g

cricket20 [7]

Hey there!:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

5 moles O2 ------------- 4 moles CO2

3.00 moles O2 ---------- ( moles of CO2 ?? )

moles of CO2 = 3.00 * 4 / 5

moles of CO2 = 12 / 5

moles of CO2 = 2.4 moles

So, molar mass CO2 = 44.01 g/mol

Therefore:

1 mole CO2 -------------- 44.01 g

2.4 moles CO2 ---------- ( mass of CO2 )

mass of CO2 = 2.4 * 44.01 / 1

mass of CO2 = 106 g

Answer A

Hope that helps!

8 0

3 years ago

What is the name of John Daltons theory

Rashid [163]

Answer:

It is called the atomic theory I believe.

Explanation:

5 0

2 years ago