27.08.2020

How many molecules are there in 45.0 grams of CH ? I
How many molecules are there in 150.0 grams of CzHs?

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There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry

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Chemistry
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There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry

Explanation:

Chemistry
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16.93\times 10^{23}\ \text{molecules}

Explanation:

Given

Mass of CH_4m=45\ gm

Molar mass of CH_4 is 12+4\times1=16

No of moles of CH_4=\dfrac{\text{Mass}}{\text{Molar mass}}=\frac{45}{16}

No of Molecules in 45gm of CH_4 is

\Rightarrow N_A\times\frac{45}{16}=6.022\times 10^{23}\times \frac{45}{16}\\\Rightarrow 16.93\times10^{23}\ \text{molecules}

Chemistry
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1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16  grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol. 

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.
Chemistry
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1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16  grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol. 

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.
Chemistry
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Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarity * L
mol = 0.15 * 0.03 = 0.0045 mol
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarit
Chemistry
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Answer: chalcogens.
Explanation: Strontium is an alkaline earth metal, it always exhibits a degree of oxidation in its compounds +2.
Chalcogens are a group of 6 chemical elements (oxygen O, sulfur S, selenium se, tellurium te, polonium Po) that have an oxidation state of -2 => Chalcogens will combine with strontium in a ratio of 1:1.
Chemistry
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Answer: b. Fiona is correct because the diagram shows two individual simple machines.

Explanation:
A mechanical device using which we can change the direction or magnitude of force applied is known as simple machine.
For example, in the given diagram there are two individual simple machines.
The machine helps in changing the direction or magnitude of force applied by the man. As a result, it becomes easy for him to carry different things easily from one place to another.
Thus, we can conclude that the statement Fiona is correct because the diagram shows two individual simple machines, is correct.
Answer: b. Fiona is correct because the diagram shows two individual simple machines.

Explanation
Chemistry
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Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
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Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the
Chemistry
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glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

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