What is the molarity of 2.0 moles of glucose
in 4.0 L of glucose solution?​

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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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]

Answer:

Because of time zones?

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

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:  

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:

I believe the answer is Reflecting Telescope

Explanation:

-leeknowishawt

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: 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.

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:

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:

H2CO

Explanation:

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:

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 %

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:

i dunno

Explanation:

Answer:

Explanation:

Evidence for early forms of life comes from fossils. By studying fossils, scientists can learn how much (or how little) organisms have changed as life developed on Earth. There are gaps in the fossil record because many early forms of life were soft-bodied, which means that they have left few traces behind.

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:

4

Explanation:

Because of single replacement the ideal shape comes in form

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:

See explanation

Explanation:

A reducing agent in chemistry is any substance that accepts electrons and its oxidation number is decreased in a redox reaction.

Manganese occurs in a variety of oxidation states; +7, +6, +5, +4, +3, +2. This makes it an exceptionally good reducing agent since it can accept different number of electrons and change from one oxidation state to another.

The commonest compound of manganese used as a strong oxidizing agent is KMnO4 in which manganese has an oxidation number of +7.

Answer:

8H2SO4 + 2KMnO4 + 5Na2O2 => 8H2O + 2MnSO4 + 5Na2SO4 + 5O2 + K2SO4

have:

Explanation:

Answer:

Explanation:

The [H3O+] of HNO2 can be found by using the formula

[tex][H_3O^{ + } ] = \sqrt{Kac} [/tex]

Ka is the acid dissociation constant

c is the concentration of the acid

From the question we have

[tex][H3O^{ + } ] = \sqrt{4.46 \times {10}^{ - 4} \times 0.1} \\ = 0.006678...[/tex]

We have the final answer as

Hope this helps you

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:

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:

A heterogeneous catalyst can be easily separated from reactants.

Explanation:

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:

d

Explanation:

According to law of conservation of mass, mass  is conserved during burning of paper although the mass of ash is less than paper as some of the products are  given out as gas.

According to law of conservation of mass, it is evident that mass is neither created nor destroyed rather it is restored at the end of a chemical reaction .

Law of conservation of mass and energy are related as mass and energy are directly proportional which is indicated by the equation E=mc².Concept of conservation of mass is widely used in field of chemistry, fluid dynamics.

Law needs to be modified in accordance with laws of quantum mechanics under the principle of mass and energy equivalence.This law was proposed by Antoine Lavoisier in the year 1789.

Learn more about law of conservation of mass ,here:

https://brainly.com/question/13383562

#SPJ5

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:

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