Base weight is a system of comparing atomic weights to carbon.

Match the characteristics of each of the substance to their category

hjlf

Answer:

Metal: C, F, G,

Nonmetal: B, E,

Metalloid: A, D, H

Explanation:

6 0

3 years ago

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t

Serga [27]

Explanation:

1)

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

2)

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0

3 years ago

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co

DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: $\frac{289kg}{1249 kg}$ =0,231 kg O₂/ kg air

I hope it helps!

4 0

2 years ago

Which of the following is an important application of hydrogen bonding

Andrews [41]

Hydrogen bonding is important because it is crucial to all life on Earth. Here are three reasons why hydrogen bonding is important. DNA has a double-helix structure because hydrogen bonds hold together the base pairs in the middle. Without hydrogen bonds, DNA would have to exist as a different structure.

6 0

3 years ago

Read 2 more answers

Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give

Ghella [55]

Answer:

$[base]=0.28M$