What is normality in chemistry?​

Answer:

True

Explanation:

The reaction between Diisopropyl ether and concentrated aqueous HI forms two initial organic products as shown in the image attached.

The hydrogen of the HI becomes attached to the oxygen in the ether leading to a cleavage of the C-O bond to yield the first compound. The I^- become attached to the other moiety in the original molecule to yield the second compound as shown in the image attached.

This question is incomplete, the complete question is;

Dimethyl ether, a useful organic solvent, is prepared in two steps.

In the first step, carbon dioxide and hydrogen react to form methanol and water:  

CO₂(g) + 3H₂(g)  →  CH₃OH(l) + H₂O(l)                ΔH₁ = -131.kJ

In the second step, methanol reacts to form dimethyl ether and water:

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                ΔH₂ = 8.kJ

Calculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions. Round your answer to the nearest kJ.

Answer:

the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ

Explanation:

Given the data in the question;

For the First Step;

CO₂(g) + 3H₂(g)  →  CH₃OH(l) + H₂O(l)                ΔH₁ = -131.kJ

For the First Step;

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                ΔH₂ = 8.kJ

Using Hess's Law of Constant Heat Summation;

" regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes "

we multiply step 1 reaction by the coefficient of 2

2CO₂(g) + 2×3H₂(g)  →  2CH₃OH(l) + 2H₂O(l)                ΔH₁ = 2 × -131.kJ

we have

2CO₂(g) + 6H₂(g)  →  2CH₃OH(l) + 2H₂O(l)                ΔH₁ = -262 kJ

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                       ΔH₂ = 8 kJ

{ 2CH₃OH cancels 2CH₃OH }

Hence, we have;

2CO₂ + 6H₂  →  CH₃OCH₃(g) + 3H₂O(l)

So According to Hess's Law;

ΔH[tex]_{sum[/tex] = ΔH₁ + ΔH₂

we substitute

ΔH[tex]_{sum[/tex] = -262 kJ + 8 kJ

ΔH[tex]_{sum[/tex] = -254 kJ

Therefore, the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ

Answer:

Explanation:

We are to match each land resource to what they are being used for.

Clay             →→→  used to make pottery and tiles

iron ore       →→→  used to make steel

Salt              →→→   used as a flavoring in food

aggregate   →→→   used in construction

graphite       →→→ used to make batteries

Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.

Iron ore:  The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.

Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.

Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,

Answer:

cis

Explanation:

Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.

Answer:

See detailed explanation.

Explanation:

Hello there!

In this case, for the given scenario, we will proceed as follows:

1. Here, we infer that the products are iron (III) hydroxide (precipitate) and calcium chloride:

[tex]3Ca(OH)_2+2FeCl_3\rightarrow 3CaCl_2+2Fe(OH)_3[/tex]

2. In this step we firstly calculate the moles of both reactants, by using their molar masses 74.093 and 162.2 g/mol respectively:

[tex]14.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093gCa(OH)_2}=0.189molCa(OH)_2 \\\\17.0gFeCl_3*\frac{1molFeCl_3}{162.2gFeCl_3}=0.105molFeCl_3[/tex]

Now, we calculate the moles of calcium hydroxide consumed by 0.105 moles of iron (III) chloride by using the 3:2 mole ratio between them:

[tex]0.105molFeCl_3*\frac{3molCa(OH)_2}{2molFeCl_3} =0.157molCa(OH)_2[/tex]

Thus, we infer that calcium hydroxide is in excess as 0.189 moles are available for it but just 0.157 moles react and therefore, iron (III) chloride is the limiting reactant.

3. Here, we use the moles of iron (III) chloride we've just computed, the 2:2 mole ratio with iron (III) hydroxide and its molar mass (106.867 g/mol) as shown below:

[tex]0.105molFeCl_3*\frac{2molFe(OH)_3}{2molFeCl_3} *\frac{106.867gFe(OH)_3}{1molFe(OH)_3} \\\\=11.2gFe(OH)_3[/tex]

Regards!

Explanation:

here's the answer to your question

Answer:

c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.

Answer:

A. Reactants change into products

Answer:

Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.

Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.

Explanation:

Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

Answer:

7 mol CuO

0.5 mol O₂

Explanation:

Step 1: Write the balanced equation

2 Cu + O₂ ⇒ 2 CuO

Step 2: Identify the limiting reactant

The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.

The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.

Since TMR > EMR, Cu is the limiting reactant

Step 3: Calculate the amount of CuO produced

7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO

Step 4: Calculate the excess of O₂ that remains

The amount of O₂ that reacts is:

7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂

The excess of O₂ that remains is:

4 mol - 3.5 mol = 0.5 mol

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Answer:

molecules of real gases occupy no volume.

Explanation:

Answer:

46.2%

Explanation:

Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles

Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.

Hence;

Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g

% yield = actual yield/theoretical yield × 100

% yield = 1.7 g/3.68 g × 100

% yield = 46.2%

Answer:

-376 kJ

Explanation:

The first step equation:

[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex]    ---- (1)

The second step equation:

[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex]      ---- (2)

To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.

From the above equations; let multiply equation (1) by 1 and equation (2) by 2.

[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex]     ---- (3)

[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex]      ----- (4)

adding the above two equations, we have:

[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]

Now, from the recent equation, we have:

2 moles of nitric acid = -752 kJ

∴

1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles

1 mole of nitric acid will be: = -376 kJ

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 1.2 M

Molarity of diluted solution (M₂) = 0.5 M

Volume of diluted solution (V₂) = 100 mL

The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:

1.2 × V₁ = 0.5 × 100

1.2 × V₁ = 50

Divide both side by 1.2

V₁ = 50 / 1.2

Thus, 42 mL of the stock solution is needed.

Learn more: https://brainly.com/question/24219233

Answer:

They need 41.7 mL of the original stock solution.

Explanation:

We can use the following equation for dilutions:

Cc x Vc = Cd x Vd

Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.

The concentrated solution is the original stock solution, and it has:

Cc = 1.2 M

The diluted solution must be:

Cd = 0.500 M

Vd = 100 mL

So, we have to calculate Vc. For this, we replace the data in the equation:

[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]

Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100  mL of a diluted solution with a concentration of 0.500 M.

Answer:

39.29 mg

Explanation:

Step 1: Calculate the partial pressure of hydrogen

The pressure of the atmosphere is equal to the sum of the partial pressures of the water and the hydrogen. (1 Torr = 1 mmHg)

P = pH₂O + pH₂

pH₂ = P - pH₂O = 758.3 mmHg - 16.5 mmHg = 741.8 mmHg

We will convert it using the conversion factor 1 atm = 760.0 mmHg.

741.8 mmHg × 1 atm/760.0 mmHg = 0.9761 atm

Step 2: Convert 19.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 19.0 + 273.15 = 292.2 K

Step 3: Calculate the mass of hydrogen

First, we will calculate the moles of hydrogen using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 0.9761 atm × 0.47910 L / (0.08206 atm.L/mol.K) × 292.2 K = 0.01950 mol

The molar mass of hydrogen is 2.015 g/mol. The mass of hydrogen is:

0.01950 mol × 2.015 g/mol = 0.03929 g = 39.29 mg

Answer:

The new temperature of the gas is 746 K.

Explanation:

Given that,

The volume of the gas, V₁ = 117 mL

Temperature, T₁ = 100°C = 373

Final volume of the gas, V₂ = 234 mL

We need to find the final temperature. The relation between temperature and volume is given by :

[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]

So, the new temperature of the gas is 746 K.

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

https://brainly.com/question/24310467

Answer:

d. End product is that product with a ketone and carboxylic acid.

Explanation:

[tex]{ \sf{NaBH_{4} : }}[/tex]

Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.

[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]

Then acidified water is an oxidising mixture which reverses the reduction reaction.

Explanation:

Option D is your answer

Hope it helps

Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

Answer:

See explanation

Explanation:

We must first write the equation of the reaction as follows;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Now;

We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles

So;

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

So;

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

We can now decide on the limiting reactant to be C3H8

Therefore;

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

Again;

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Hence;

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

In order to obtain the percentage yield from the reaction, we have;

b) Actual yield = 269.34 g

Theoretical yield = 396 g

Therefore;

% yield = actual yield/theoretical yield × 100/1

Substituting values

% yield = 269.34 g /396 g × 100

% yield = 68%

Answer:

Alpha and beta particles originate from the nucleus, protection from radiation is important because if the radiation passes through the body it can damage cells. Exposure to radiation is often limited by increasing the distance from the radioactive source.

Explanation:  

Answer:

See the answer below

Explanation:

From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.

Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.

Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.

In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.

Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

Answer:

Specific gravity of the saturated solution is 2

Explanation:

The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:

Specific gravity  = Density

The density is the ratio between the mass of the solution and its volume.

In 100mL of water, the mass of KI that can be dissolved is:

100mL * (1g KI / 0.7mL) = 143g of KI

That means all the 100g of KI are dissolved (Mass solute)

As the volume of water is 100mL, the mass is 100g (Mass solvent)

The mass of the solution is 100g + 100g = 200g

In a volume of 100mL, the density of the solution is:

200g / 100mL = 2g/mL.

The specific gravity has no units, that means specific gravity of the saturated solution is 2