A bio catalyst that increases the rate of the reaction without being changed
a) Aluminum oxide. b) Silicon dioxide. c) Enzyme. d) Hydrogen peroxide
43. What is the
than the reaction substrate.
42. A

Answer:

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Explanation:

even i know how to type şïllily

Answer:

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Explanation:

Let's consider the balanced chemical equation that takes place when sulfuric acid solution and solid aluminum hydroxide react to form aluminum sulfate and water. This is a neutralization equation.

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Answer:

Acids react with most metals to form hydrogen gas and salt. ... When an acid reacts with metal, salt and hydrogen gas are produced

Answer:

a) Hence, T = 207 K.

b) Hence, T2 = 226 K.

Explanation:

Now the given,

n = 0.27 moles ; P = 2.5 atm ; T = 298 K

a) γ = 5/3 since Ne is a monoatomic gas.

[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]

Hence, T = 207 K

b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]

[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]

But P = P2

[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]

This gives us:

[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]

Hence, T2 = 226 K

Answer:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Explanation:

The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:

In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).

The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Answer:

Explanation:

a) Ionic

Lithium oxide

b) Covalent

[tex]$\ce{N_2O_3}$[/tex]

c) Covalent

Phosphorus trichloride

d) Ionic

[tex]Mn_2O_3[/tex]

Answer:

(d) the atoms it contains and the way these atoms are connected

Explanation:

hope it will be helpful for you

Explanation:

Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.

Answer:

1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Explanation:

Let's consider the dissolving equations for the following compounds.

1. Rubidium hydroxide

RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Sodium carbonate

Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. Ammonium selenite

(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Answer:

.87

Explanation:

p = m/V

43.5/50

.87

Answer:

56.8%

Explanation:

The reaction of the problem is 1:1. That means 1 mole of 2-methylcyclohexanone produce 1 mole of 2-methylcyclohexanol.

Percentage yield is defined as 100 times the ratio between actual yield of the reaction (5.14g) and the theoretical yield.

The theoretical yield (All reactant produce products) is obtained from the mass of the reactant as follows:

Theoretical Yield:

8.89g 2-methylcyclohexanone * (1mol/112.17g) = 0.07925 moles 2-methylcyclohexanone

Assuming all reactant produce the product in a 100% of yield, the moles of 2-methylcyclohexanol are 0.07925 moles and the mass (Theoretical yield) is:

0.07925 moles 2-methylcyclohexanol * (114.19g/mol) = 9.05g

Percentage yield:

5.14g / 9.05g * 100 = 56.8%

The percentage of the mass successfully converted into a new product is 57.2%.

The mass of the reactant (2-methylcyclohexanone) before the reduction is given as 8.89 g.

The mass of the product ( 2-methylcyclohexanol) produced is given as 5.14 g.

The percentage of the mass successfully converted into a new product is calculated as follows;

[tex]= \frac{5.14}{8.99} \times 100\% \\\\= 57.2 \ \%[/tex]

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Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

Answer:

The correct answer is C. Nucleic acid

Explanation:

Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:

- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA

- phosphate group

- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.

Answer:

[tex]Q=292240.8J=292.2kJ[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the general heat equation:

[tex]Q=mC(T_2-T_1)[/tex]

For us to plug the given mass, specific heat and temperature change to obtain the required heat:

[tex]Q=120g*32.91\frac{J}{g*K} (98\°C-24\°C)\\\\Q=292240.8J=292.2kJ[/tex]

Regards!

D.22

is my answer than welcome

Answer: The balanced equation for the complete oxidation reaction of ethanol is [tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

Explanation:

Combustion is the chemical process where an organic molecule reacts with oxygen gas present in the air to produce carbon dioxide and water molecules.

It is also known as an oxidation reaction because oxygen is getting added.

The chemical equation for the oxidation of ethanol follows:

[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

By stoichiometry of the reaction:

1 mole of ethanol reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 3 moles of water molecules.

Answer:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Double replacement reaction.

It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Explanation:

Hello there!

In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Regards!

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

Explanation:

Answer:

the answer is d.) potassium

Explanation:

Explanation:

The term "basic unit of matter "refers to atom

A Atom

Answer:

irreversible.

I hope this will help you

Answer:

Density of liquid = 1.59 g/cm³

Explanation:

We'll begin by converting 0.610 L to cm³. This can be obtained as follow:

1 L = 1000 cm³

Therefore,

0.610 L = 0.610 L × 1000 cm³ / 1 L

0.610 L = 610 cm³

Finally, we shall determine the density of the liquid. This can be obtained as follow:

Volume of liquid = 610 cm³

Mass of liquid = 972 g

Density of liquid =?

Density = mass / volume

Density of liquid = 972 / 610

Density of liquid = 1.59 g/cm³

The density of any given liquid is equivalent to the mass of the liquid divided by the volume of the liquid.

From the given information, we have:

The mass of the unknown liquid to be = 972 g

The volume of the unknown liquid to be = 0.610 L = 610 cm³

If the formula for calculating [tex]\mathbf{Density = \dfrac{Mass}{Volume}}[/tex]

Then;

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = \dfrac{972 \\ g}{610 \ cm}}[/tex]

[tex]\mathbf{Density \ of \ the \ unknown \ liquid = 1.593442623 \ g/cm}[/tex]

The Density of the unknown liquid ≅ 1.593  g/cm³

In conclusion, the density of the unknown liquid is 1.593  g/cm³ to 3 significant figures.

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Answer:

its kcl ,option.b)

Explanation:

because its contains k+and cl- which is and ionic bond

Answer:

19.3 g of NaCl can be produced

Explanation:

We state the reaction:

Cl₂ (g) + 2NaI (s) →  2NaCl (s) + I₂ (s)

We need to determine limiting reagent:

15.4 g . 1mol /70.9g = 0.217 moles of chlorine

49.6 g . 1mol / 149.89g = 0.331 moles of NaI

Ratio is 1:2. 1 mol of chlorine reacts to 2 moles of NaI

0.217 moles may react to (0.217 . 2)/1 = 0.434 moles of NaI

It is ok to say the NaI is the limting reactant because we need 0.434 moles of it and we only have 0.331.

Ratio is 2:2.

0.331 moles of NaI can produce 0.331 moles of NaCl

We convert mass to moles: 0.331 mol . 58.45g /mol = 19.3 g

In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.

Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase

Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.

The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids

Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

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Answer:

Explanation:

We can calculate the Rf values by using the following formula:

Rf = Distance from the baseline / Solvent front distance

With that in mind we now proceed to calculate the Rf value for both components:

Answer:

[tex]\boxed {\boxed {\sf 8.22 \ \textdegree C}}[/tex]

Explanation:

The question asks us to calculate the temperature of the gas at the higher volume. Since the pressure is constant, we are only concerned about volume and temperature. We will use Charles's Law. This states that the volume of a gas and the temperature of the gas have a directly proportionate relationship. The formula is:

[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]

The gas starts at a volume of 4.17 liters and a temperature of 7.62 degrees Celsius.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{V_2}{T_2}[/tex]

The gas is expanded to a new volume of 4.50 liters, but the temperature is unknown.

[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{4.50 \ L}{T_2}[/tex]

We want to solve for the temperature at a higher volume. We must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

[tex]4.17 \ L * T_2 = 4.50 \ L * 7.62 \textdegree C[/tex]

The variable is being multiplied by 4.17 liters. The inverse of multiplication is division. Divide both sides by 4.17 L.

[tex]\frac {4.17 \ L * T_2 }{4.17 \ L}= \frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

[tex]T_2=\frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]

The units of liters (L) cancel.

[tex]T_2=\frac{4.50 * 7.62 \textdegree C}{4.17 }[/tex]

[tex]T_2=\frac{34.29}{4.17 } \textdegree C[/tex]

[tex]T_2=8.22302158273 \textdegree C[/tex]

The original measurements of liters and temperature have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

The 3 to the right in the thousandth place tells us to leave the 2 in the hundredth place.

[tex]T_2 \approx 8.22 \ \textdegree C[/tex]

The temperature of the gas at the higher volume is approximately 8.22 degrees Celsius.

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .