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Arte-miy333 [17]
3 years ago
7

Which of the following is NOT a symbol for an element? H Li Pb HF

Chemistry
2 answers:
sp2606 [1]3 years ago
3 0
HF because an element is always a capital then a lowercase letter
ohaa [14]3 years ago
3 0
HF because an element always has a capital letter first then a lowercase.
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Write the equilibrium constant expression, Kc, for the following reaction: If either the numerator or denominator is 1, please e
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Answer:

Explanation:

THE QUESTION SHOULD BE

Write the equilibrium constant expression, Kc, for the following reaction: If either the numerator or denominator is 1, please enter 1 CaCO3(s).

CaCO3(s) <=> Ca2+(aq) + CO32-(aq)

Write the mathematical form of the concentration of the products, then divide it by the concentration of the reactants as follows

Kc =[Ca2+]^1 . [CO32-]^1 / [CaCO3]^1

5 0
3 years ago
7. How many liters of NH3, at STP will react with 10.6 g O2 to form NO2 and water? 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) *
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7) Answer is: c. 4.24 L.

Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

m(O₂) = 10.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 10.6 g ÷ 32 g/mol.

n(O₂) = 0.33 mol; amount of oxygen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

V(O₂) = n(O₂) · Vm.

V(O₂) = 0.33 mol · 22.4 L/mol.

V(O₂) = 7.42 L; volume of oxygen.

From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.

n(NH₃) = 4 · 0.33 mol ÷ 7.

n(NH₃) = 0.188 mol; amount of ammonia.

V(NH₃) = 0.188 mol · 22.4 L/mol.

V(NH₃) = 4.24 L.

8) Answer is: a. 3.7 g.

Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).

m(P₄) = 3.4 g; mass of phosphorous.

n(P₄) = m(P₄) ÷ M(P₄).

n(P₄) = 3.4 g ÷ 123.9 g/mol.

n(P₄) = 0.0274 mol; limiting reactant.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.

n(PH₃) = 4 · 0.0274 mol.

n(PH₃) = 0.1096 mol.

m(PH₃) = 0.1096 mol · 34 g/mol.

m(PH₃) = 3.726 g.

9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.

Percent yield = actual yield / theoretical yield.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

For example:

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

3 0
3 years ago
Human bone does not contain?
tia_tia [17]
Human bone does not contain oxygen
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A balloon (approximate as a sphere) is inflated such that its
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Answer: 1436 mL

Explanation: 4/3 • 3.14 • 7^3 <— calculate

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
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