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2 years ago

How do you define medium and high level profession?according to class 8?

Chemistry

1 answer:

abruzzese [7]2 years ago

3 0

Answer:

Answer: medium level profession can be defined as some one who has moved past entry level but isn't nearing the end . whereas high level profession means a person has moved past entry level and also has reached the end.

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Sliver tarnishes physical or chemical change ?

elena55 [62]

silver tarnishes is a chemical change

5 0

3 years ago

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How many grams are in 4.5 x 10^22 molecules of Ba(NO2)2

NNADVOKAT [17]

Answer:

17 g Ba(NO₂)₂

General Formulas and Concepts:

Chemistry

Stoichiometry
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

4.5 × 10²² molecules Ba(NO₂)₂

Step 2: Define conversion

Molar Mass of Ba - 137.33 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(NO₂)₂ - 137.33 + 2(14.01) + 4(16.00) = 229.35 g/mol

Step 3: Dimensional Analysis

$4.5 \cdot 10^{22} \ mc \ Ba(NO_2)_2(\frac{1 \ mol \ Ba(NO_2)_2}{6.022 \cdot 10^{23} \ mc \ Ba(NO_2)_2} )(\frac{229.35 \ g \ Ba(NO_2)_2}{1 \ mol \ Ba(NO_2)_2} )$

= 17.1384 g Ba(NO₂)₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

17.1384 g Ba(NO₂)₂ ≈ 17 g Ba(NO₂)₂

7 0

2 years ago

What is chromatography usese of chromatography ?

victus00 [196]

Explanation:

chromatoghraphy is a laboratory technique for the separation of mixture... ...

chromatoghraphy may be preparative or analytical. The purpose of preparative chromatoghraphy is to separate the components of a mixture for late use, ans is thus a form of purification..... I is your answer it may help you

5 0

2 years ago

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A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

alexandr402 [8]

Answer:

$x_{N_2F_2}= 0.415\\\\x_{SF_6}=0.585$

Explanation:

Hello!

In this case, since the mole fraction of both gases in the tank is computed via:

$x_{N_2F_2}=\frac{n_{N_2F_2}}{n_{N_2F_2}+n_{SF_6}} \\\\x_{SF_6}=\frac{n_{SF_6}}{n_{N_2F_2}+n_{SF_6}}$

It means we need to compute the moles of each gas, just as it is shown down below:

$n_{N_2F_2}}=5.53gN_2F_2*\frac{1molN_2F_2}{66.01gN_2F_2} =0.0838molN_2F_2\\\\n_{SF_6}=17.3gSF_6*\frac{1molSF_6}{146.06gSF_6} =0.118molSF_6$

Thus, the mole fractions turn out:

$x_{N_2F_2}=\frac{0.0838mol}{0.0838mol+0.118mol}= 0.415\\\\x_{SF_6}=\frac{0.0838mol}{0.0838mol+0.0838mol}=0.585$

Best regards!

8 0

2 years ago

2AlF3 + 3K2O → 6KF + Al2O3 How many grams of AlF3would it take to make 15.524 g of KF?

mr_godi [17]

<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table

<h3>Explanation:</h3>

Step 1: Define

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

Step 2: Identify Conversions

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0

2 years ago

Read 2 more answers