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Vedmedyk [2.9K]
3 years ago
8

In the conclusion to the snake mole lab, you have to convert 3.45 moles of (NH4)2S04 to

Chemistry
1 answer:
Serhud [2]3 years ago
3 0
Molar mass = 132.14g/mol
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High-pressure liquid chromatography (HPLC) is a method used in chemistry and biochemistry to
Alchen [17]

Answer:

2,400,000 torr (3 s.f.)

Explanation:

Convert the pressure from Pascal to atm first:

\boxed{1 \: atm = 101325 \:Pa }

3.20 ×10⁸ Pa

= [(3.20 ×10⁸) ÷101325] atm

= 3158.2 atm (5 s.f.)

Convert atm to torr:

\boxed{1 \: atm = 760 \: torr}

3158.2 atm

= (3158.2 ×760) torr

= 2400000 torr (3 s.f.)

6 0
2 years ago
Which product is typically made using softwood?
deff fn [24]

Answer:c

Explanation:

softwood is used in doors roofs and so on

5 0
3 years ago
Read 2 more answers
How many grams of ammonium chloride (gram formula mass= 53.5 g) are contained in .500 L of a 2.00 M solution?
Elza [17]

Answer:

53.5g of NH4Cl

Explanation:

First, we need to obtain the number of mole of NH4Cl. This is illustrated below:

Volume = 0.5L

Molarity = 2M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 2 x 0.5

Mole = 1mole

Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:

Molar Mass of NH4Cl = 53.5g/mol

Number of mole = 1

Mass =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass = 1 x 53.5

Mass = 53.5g

Therefore, 53.5g of NH4Cl is contained in the solution.

8 0
3 years ago
1) What is the electron configuration of the element with 27 protons?
Romashka [77]
1) cobalt u can tell bc on a periodic table there is a small number that on cobalt is 27.

2) think that would be 11 bc in the 4th shell there can  be up to 18 electrons
6 0
3 years ago
in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,
insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, H_{supplied} = 3.34 × 10³ J

The amount thermal energy gained by the kettle, H_{absorbed}  = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}}  \times 100

Therefore, we get;

\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3}  \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0
3 years ago
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