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3 years ago

HELPPPPPPP

Chemistry

2 answers:

LenKa [72]3 years ago

7 0

B. G—T—A—G—C—T
(Ive answered this

marusya05 [52]3 years ago

5 0

Which nitrogen base sequence is the partner of C-A-T-C-G-A?
b. G-T-A-G-C-T

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In an experiment, a compound was determined to contain 68.94% oxygen and 31.06% of an unknown element by weight. The molecular w

monitta

Answer is: the compound is B₂O₃.

ω(O) = 68.94% ÷ 100%.

ω(O) = 0.6894; percentage of oxygen in the compound.

ω(X) = 31.06% ÷ 100%.

ω(X) = 0.3106; percentage of unknown element in the compound.

If we take 69.7 grams of the compound:

M(compound) = 69.7 g/mol.

n(compound) = 69.7 g ÷ 69.7 g/mol.

n(compound) = 1 mol.

n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.

n(O) = 3 mol.

M(compound) = n(O) · M(O) + n(X) · M(X).

n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.

n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).

8 0

3 years ago

What is the percentage composition for H2O in this compound Al(OH)3.2H2O?

Gelneren [198K]

Answer:

32.73%

Explanation:

To solve this problem, first find the molar mass of Al(OH)₃.2H₂O

Atomic mass of Al = 27g/mol

O = 16g/mol

H = 1g/mol

Molar mass = 27 + 3(16 + 1) + 2(2(1) + 16)

= 27 + 51 + 36

= 114g/mol

Percentage composition = $\frac{36}{114}$ x 100 = 32.73%

6 0

3 years ago

Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2

Usimov [2.4K]

Answer:

$\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 98.08 392.18

2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

Use the molar mass of H₂SO₄ to convert the mass of H₂SO₄ to moles of H₂SO₄
Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

$\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}$

(ii) Moles of H₂SO₄

$\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}$

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ $\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}$

(iv) Mass of Cr₂(SO₄)₃ $\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}$

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

$\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}$

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4 years ago

The ____ allows you to compare the elements and understand their properties

sweet-ann [11.9K]

Answer:

Periodic Table

hope this helps

have a good day :)

Explanation:

5 0

3 years ago

Read 2 more answers

Is Lithium oxide type I or type II ?

AnnZ [28]

Answer:

1 is the answer

Explanation:

The ionic formula for Lithium Oxide is Li2O . Lithium is an Alkali Metal in the first column of the periodic table. This means that lithium has 1 valence electron it readily gives away in order to seek the stability of the octet.

8 0

3 years ago