write down any four uses of solution






.I will mark him\her as brilliant

Answer:

4 Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Explanation:

Let's consider the unbalanced molecular equation when aluminum metal and oxygen gas combine to produce aluminum oxide. This is a combination reaction.

Al(s) + O₂(g) ⇒ Al₂O₃(s)

We can balance O  atoms by multiplying O₂ by 3 and Al₂O₃ by 2.

Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Finally, we multiply Al by 4 to get the balanced equation.

4 Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Answer:

C               (I think)

Explanation:

the levels are higher than they have ever been

Answer:

ca you attach these two pdf ?? or provide the link to this document

Answer:

M = 1.04 M

Explanation:

Given data:

Molarity of solution = ?

Mass of HCl = 6.27 g

Volume of solution = 163 mL (163 mL× 1L /1000 mL = 0.163 L)

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 6.27 g / 36.5 g/mol

Number of moles = 0.17 mol

Molarity:

M = 0.17 mol/ 0.163 L

M = 1.04 M

Answer:

4.6 M HCI

Explanation:

on edge

Answer:

0.98 M

Explanation:

Given data:

Mass of sugar = 20 g

Mass of water = 59 g

Concentration of solution = ?

Solution:

1 g = 0.01 L

59 g× 0.01 L/1 g

0.059 L

Concentration formula:

C = m/v

m = moles of solute

v = volume of solution

Number of moles of sugar:

Number of moles = mass/molar mass

Number of moles = 20 g/ 342.3 g/mol

Number of moles = 0.058 mol

Now we will calculate the concentration:

C = 0.058 mol/0.059 L

C = 0.98 mol/L

C = 0.98 M

Answer:

0.00752L of 0.305M K₃PO₄

Explanation:

Based on the balanced chemical equation, 2 moles of K₃PO₄ react with 3 moles of NiCl₂.

To find the volume of K₃PO₄ we need to determine the moles of this compound required for a complete reaction finding moles of NiCl₂ and using the chemical equation:

Moles NiCl₂:

187mL = 0.187L * (0.0184mol / L) = 0.00344 moles NiCl₂

Moles K₃PO₄:

As 2 moles of K₃PO₄ react with 3 moles of NiCl₂:

0.00344 moles NiCl₂ * (2 mol K₃PO₄ / 3 moles NiCl₂) =

0.002294 moles of K₃PO₄

Volume of 0.305M K₃PO₄:

0.002294 moles of K₃PO₄ * (1L / 0.305moles K₃PO₄) =

Answer:

Approximately [tex]0.224\;\rm L[/tex], assuming that this reaction took place under standard temperature and pressure, and that [tex]\rm CO_2[/tex] behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that [tex]\rm CaCO_3[/tex] is the limiting reactant. How many moles of [tex]\rm CO_2[/tex] would be produced?

Look up the relative atomic mass of [tex]\rm Ca[/tex], [tex]\rm C[/tex], and [tex]\rm O[/tex] on a modern periodic table:

Calculate the formula mass of [tex]\rm CaCO_3[/tex]:

[tex]\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

Calculate the number of moles of formula units in [tex]1\; \rm g[/tex] of [tex]\rm CaCO_3[/tex] using its formula mass:

[tex]\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}[/tex].

In the balanced chemical equation, the ratio between the coefficient of [tex]\rm CaCO_3[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1[/tex].

In other words, for each mole of [tex]\rm CaCO_3[/tex] formula units consumed, one mole of [tex]\rm CO_2[/tex] would be produced.

If [tex]\rm CaCO_3[/tex] is indeed the limiting reactant, all that approximately [tex]1.00\times 10^{-2}\; \rm mol[/tex] of [tex]\rm CaCO_3\![/tex] formula would be consumed. That would produce approximately [tex]1.00\times 10^{-2}\; \rm mol\![/tex] of [tex]\rm CO_2[/tex].

On the other hand, assume that [tex]\rm HCl[/tex] is the limiting reactant.

Convert the volume of [tex]\rm HCl[/tex] to [tex]\rm dm^{3}[/tex] (so as to match the unit of concentration.)

[tex]\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}[/tex].

Calculate the number of moles of [tex]\rm HCl[/tex] molecules in that [tex]5.00\times 10^{-2}\; \rm dm^{3}[/tex] of this [tex]\rm 0.05\; \rm mol \cdot dm^{-3}[/tex]

[tex]\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}[/tex].

Notice that in the balanced chemical reaction, the ratio between the coefficient of [tex]\rm HCl[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}[/tex].

In other words, each mole of [tex]\rm HCl[/tex] molecules consumed would produce only [tex]0.5\;\rm mol[/tex] of [tex]\rm CO_2[/tex] molecules.

Therefore, if [tex]\rm HCl[/tex] is the limiting reactant, that [tex]2.50 \times 10^{-3}\; \rm mol[/tex] of [tex]\rm HCl\![/tex] molecules would produce only one-half as many (that is, [tex]1.25\times 10^{-3}\; \rm mol[/tex]) of [tex]\rm CO_2[/tex] molecules.

If [tex]\rm CaCO_3[/tex] is the limiting reactant, [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. However, if [tex]\rm HCl[/tex] is the limiting reactant, [tex]1.25\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be produced.

In reality, no more than [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. The reason is that all [tex]\rm CaCO_3[/tex] would have been consumed before [tex]\rm HCl[/tex] was.

After finding the limiting reactant, approximate the volume of the [tex]\rm CO_2\![/tex] produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately [tex]22.4\; \rm L[/tex].

If [tex]\rm CO_2[/tex] behaves like an ideal gas, the volume of that [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be approximately [tex]\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L[/tex].

Answer:  

3.2 moles of carbon dioxide

Explanation:

When C₂H₂ reacts with O₂ the chemical reaction is:

C₂H₂ + 5/2O₂ → 2CO₂ + H₂O

Where 1 mole of acetylene produce 2 moles of carbon dioxide.

Based on the chemical equation of this reaction, there are produced twice the moles of acetylene in moles of carbon dioxide assuming a compete reaction.

If 1.6 moles of acetylene reacts there are produced:

1.6 moles Acetylene * (2 moles Carbon Dioxide / 1 mole Acetylene) =

3.2 moles of carbon dioxide

Answer:

Because of time zones?

Explanation: The only thing that would make sense to me is the time zones.

Answer:

52.4%

Explanation:

Since a simple cubic unit cell contains only 1 atom. The packing efficiency of the simple cubic cell is 52.4 %

Answer:

I believe the answer is Reflecting Telescope

Explanation:

-leeknowishawt

mol fraction Codein : 0.054

mol fraction Ethanol : 0.946

Given

46.85 g of codeine

125.5 g of ethanol

Required

mol fraction

Solution

The mole fraction : the mole ratio of a substance to the mole of solution /mixture  

mol Codeine, C₁₈H₂₁NO₃(MW=299.4 g/mol) :

[tex]\tt mol=\dfrac{46.85}{299.4 g/mol}\\\\mol=0.1565[/tex]

mol Ethanol, C₂H₅OH(MW=46.07 g/mol)

[tex]\tt mol=\dfrac{125.5}{46.07}=2.7241[/tex]

mol of solution :

[tex]\tt mol~codein+mol~ethanol=0.1565+2.7241=2.8806[/tex]

[tex]\tt \dfrac{0.1565}{2.8806}=0.054[/tex]

mol fraction Ethanol :

[tex]\tt \dfrac{2.7241}{2.8806}=0.946[/tex]

The oxidation number for S in the compound SO₃ : +6

The formula for determining Oxidation Numbers :

1. Single element atomic oxidation number = 0.

Group IA (Li, Na, K, Rb, Cs, and Fr): +1

Group IIA (Be, Mg, Ca, Sr and Ba): +2

H in compound = +1, except metal hydride compounds (Hydrogen which binds to groups IA or IIA) oxidation number H = -1, for example, LiH, MgH₂, etc.

2. Oxidation number O in compound = -2, except OF₂ = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.

3 The oxidation number in an uncharged compound = 0,

Total oxidation number in ion = ion charge, Example NO₃⁻ = -1

The oxidation number for S in SO₃ :

[tex]\tt oxidation~number~of~S+3\times oxidation~number~of~O=0\\\\oxidaton~number~of~S+3\times(-2)=0\\\\oxidation~number~of~S=+6[/tex]

Answer:

35.1 g

Explanation:

Given data:

Mass of Al(OH)₃ = 100 g

Mass of water formed = ?

Solution:

Chemical equation:

2Al(OH)₃      →        Al₂O₃ + 3H₂O

Number of moles of Al(OH)₃:

Number of moles = mass/molar mass

Number of moles = 100 g/ 78 g/mol

Number of moles = 1.3 mol

now we will compare the moles of Al(OH)₃ with water.

                          Al(OH)₃         :          H₂O

                             2                :            3

                             1.3              :           3/2×1.3 = 1.95

Mass of water produced:

Mass = number of moles ×molar mass

Mass = 1.95 mol × 18 g/mol

Mass = 35.1 g

Answer: Positive ions or cations

Explanation:

The ions are formed when an atom gains or lose electrons.

The ions are classified into two which are called the cations and anions. The cation is classified as the positive charge formed by the loss of electrons. Example: [tex]Na^+[/tex] is formed by the loss of one electron by [tex]Na[/tex]

The anion is classified as the negative charge ion formed by the gain of electrons. Example : [tex]F^-[/tex] is formed the gain of one electron by [tex]F[/tex]

Thus when an element loses electrons, it forms positive ions or cations.

Answer:

V₂ =  116126.75 cm³

Explanation:

Given data:

Radius of balloon = 15 cm

Initial pressure = 2 atm

Initial temperature = 35 °C (35 +273 = 308K)

Final temperature = -20°C (-20+273 = 253 K)

Final pressure = 0.3 atm

Final volume = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

Initial volume of balloon:

V = 4/3πr³

V = 4/3×22/7×(15cm)³

V = 14137.17 cm³

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2 atm × 14137.17 cm³ × 253 K / 308 K × 0.3 atm

V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm

V₂ =  116126.75 cm³

Answer:

erosion

Explanation:

erosion is the term that refers to the process by which land is worn away by natural forces or human activity.

Answer:

Erosion

Explanation:

Just took the test, hope it helps!

Answer:

Glucose is combined with oxygen and releases usable energy, carbon dioxide, and water. Cells can use that extra energy to power their functions. The energy isn't just floating around. It's stored in an excitable compound called ATP (adenosine triphosphate).

Explanation:

Answer: ATP(adenosine triphosohate) is released with water and carbon dioxide as byproducts.

Explanation: Remember that water, CO2,and ATP are the products of cellular respiration.

Answer:

The four elements most commonly found in living things are carbon, nitrogen, hydrogen, and oxygen.

Explanation:

The four elements most commonly found in living things are carbon, nitrogen, hydrogen, and oxygen.

Answer:

Chemistry can be applicable in many aspects which are

i) At school, through getting knowledge you can assess different aspect of chemistry, eg production of Chemicals.

ii) At hospital, also chemistry can be applicable in hospital because most of medicine and other materials are made up from chemistry application.

iii) Agriculture, through agricultural materials it simply because you can get more efficient through planted by using manures and fertilizer which are made up from chemistry application.

Answer:

it is 7

Explanation:

dogs are 7 HOPE THIS HELPED kidding i don get the question

Explanation:

DON"T EAT MY DOG OR I"LL EAT YOU

Answer:

A heterogeneous catalyst can be easily separated from reactants.

Explanation:

Answer:

major enantiomer = 90.5 %

minor enantiomer = 9.5 %

Explanation:

Assuming that the  major isomer is in excess;

From, ee/2 + 50

ee = enantiomeric excess = 81%

%  major =81/2 + 50 = 90.5 %

%  minor = 100% - 90.5 %

%  minor = 9.5 %

B. Sodium hydroxide and Hydrochloric acid

Chemical equations can be expressed in terms of

In a chemical reaction, there are several components, including reactants, products, and compounds that react in the form of solids, gases, and liquids

Reactants are compounds that react and form products

The reactants are generally to the left of the chemical equation and the products are to the right of the reaction

The statement of the word reaction above can be expressed by the reaction formula

[tex]\tt NaOH+HCl\rightarrow NaCl+H_2O[/tex]

This reaction is called a neutralization reaction because it involves acids(HCl) and bases(NaOH) and produces salt(NaCl)and water(H₂O)

From the reaction above, the reactants on the left are NaOH and HCl compounds

Answer:

H2CO

Explanation:

Answer:

D

Explanation:

The other three are related to

a. selling a home

b. buying a home

c. owning a home

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