Given that 0.403 g of nitrogen combines with 0.0864 g of hydrogen and H is assigned an atomic mass of 1.00 amu calculate the atomic mass of nitrogen based on the formula NH.

The answer is A,Today is cloudy, but tomorrow will be clear and sunny.

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Answer:

the mixture is made of one substance

Explanation:

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The enzyme α-amylase increases the rate of starch breakdown by lowering the activation energy of the reaction, which is option (e). Enzymes act as catalysts in metabolic processes, and their activity can be influenced by environmental factors such as pH and temperature.

The enzyme α-amylase increases the rate at which starch is broken down into smaller oligosaccharides by lowering the activation energy of the reaction. This process does not require altering the equilibrium constant or the change in free energy; rather, it involves facilitating the reaction so that the energy barrier is lower, allowing the reactants to convert into products more easily. In the options provided, the correct answer to how α-amylase increases the rate of the reaction is (e) lowering the activation energy of the reaction.

An enzyme's role is to act as a catalyst, speeding up chemical reactions without being consumed or permanently changed by the reaction. Enzymes are crucial for metabolic processes, such as the hydrolysis of starch into sugars like glucose and maltose, which are used by cells as a source of energy and carbon. Parameters that can influence the activity of α-amylase include pH and temperature, which if not optimal, can reduce the rate of the reaction.

An element existing in nature by itself, such as a neutral atom, will have an overall charge of zero due to the balanced number of protons and electrons.

When an element exists in nature by itself, it is in the form of a neutral atom, which means it must have an overall charge of zero. This neutrality is achieved because the atom has an equal number of protons and electrons, and since protons and electrons carry equal but opposite charges, their charges cancel each other out. The proton carries a positive charge, and the electron carries a negative charge, both with a magnitude of 1.60×10-19 coulombs (C). Neutral atoms may form molecules by sharing electrons via covalent bonds but still retain no net charge unless they lose or gain electrons to become ions. Therefore, in its most stable form, an element on its own will exhibit a charge of zero.

Intramolecular forces such as the ionic bond strength in a CrCl2 molecule tend to be stronger than the intermolecular forces, which include the attractive forces between water molecules and chromium chloride ions.

The attractive forces between water molecules and chromium (Cr) and chloride (Cl) ions, or the ionic bond strength in a CrCl2 molecule, can be compared by analyzing intermolecular forces and intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, like the ionic bonds in a CrCl2 molecule. These forces are typically much stronger compared to intermolecular forces. Intermolecular forces, on the other hand, are the attractions between molecules, such as the attractive forces between water and Cr and Cl ions. Using water as an example, it has strong intermolecular forces of hydrogen bonding leading to high surface tension. However, the energy required to overcome these forces (around 17 kilojoules for one mole of water) is significantly less than the energy required to break the covalent bonds in the water molecule itself (about 430 kilojoules). Therefore, intramolecular forces like the ionic bond strength of CrCl2 are generally stronger than intermolecular forces between water molecules and Cr and Cl ions.

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The partial pressure of Hydrogen gas can directly be calculated by simply taking the difference of the overall pressure and the vapour pressure of water. That is:

P (H2 gas) = 759.2 torr – 23.8 torr

P (H2 gas) = 735.4 torr

The weight percent of water in the hydrate before heating is 16.58%.

To find the weight percent of water in the hydrate before heating, we need to calculate the percent by mass of water. This can be done by dividing the mass of water in 1 mole of the hydrate by the molar mass of the hydrate, and then multiplying by 100%. The formula for weight percent is:

Given that the weight of water is 0.349 grams and the weight of the hydrate is 2.107 grams, we can substitute these values into the formula:

Hence, The weight percent of water in the hydrate before heating is 16.58%.

To find the number of moles of gas gained or lost, we can use the ideal gas law equation. By substituting the given values into the equation and simplifying, we find that the system gained or lost 0.921 moles of gas.

To find the number of moles of gas gained or lost, we will use the ideal gas law equation:

P1V1 / T1 = P2V2 / T2

Using the given information:

P1 = 1 atm, V1 = 22.4 L (at STP), T1 = 273 K

P2 = 2.00 atm, V2 = 14.5 L, T2 = 100 °C = 373 K

Substituting the values into the equation:

(1 atm)(22.4 L) / (273 K) = (2.00 atm)(14.5 L) / (373 K)

Simplifying the equation:

Moles gained or lost = (1 atm)(22.4 L)(373 K) / (273 K)(2.00 atm)(14.5 L)

Calculating the result:

Moles gained or lost = 0.921 moles

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16.34 km/2.34 h = 6.98 km/h

The product of radioactive decay :

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles

General formulas used in decay:

[tex] \large {\boxed {Nt = No. (\frac {1} {2}) ^ {\frac {T} {t1 / 2}}}} [/tex]

Information:

T = duration of decay

t1 / 2 = elemental half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

At the core, the reaction applies the law of eternity

• 1. energy

the energy before and after the reaction is the same

• 2. atomic number

the number of atomic numbers before and after the same

• 3. mass number

the number of mass numbers before and after the same

Particles that play a role in core reactions include

• alpha α particles ₂He⁴

atomic number decreases by 2, mass number decreases by 4

• beta β ₋₁e⁰ particles

atomic number remains the same , mass number decreases by -1

• gamma particles γ

Core reactions that occur usually release high energy

• positron particles ₁e⁰

atomic number remains the same , mass number decreases by +1

In general, the core reaction equation can be written:

X = target core

a = particle fired

Y = new core

b = the particle produced

Q = heat energy

or can be written simply

X (a, b) Y

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To prepare 1.80 L of a 0.215 M H2SO4 solution, you will need to use 21.5 mL of 18 M sulfuric acid.

To calculate the volume of 18 M sulfuric acid needed, we can use the formula:

M1V1 = M2V2

Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Plugging in the given values:

(0.215 M)(1.80 L) = (18 M)V2

Solving for V2:

V2 = (0.215 M)(1.80 L) / (18 M) = 0.0215 L = 21.5 mL

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To prepare 1.80 L of 0.215 M H2SO4, you would need to use 21.5 mL of 18 M sulfuric acid. This is calculated using the formula for dilution: C1V1 = C2V2.

To answer this question, you need to use the formula for dilution of solutions: C1V1 = C2V2. Here, C1 is the initial concentration(18 M), V1 is the volume of this concentrated solution that we're trying to find, C2 is the final concentration(0.215 M), and V2 is the final volume(1.80 L).

When we plug these numbers into the formula, we get the equation 18 M × V1 = 0.215 M × 1.80 L. Solving for V1 gives us V1 = (0.215 M × 1.80 L) / (18 M) = 0.0215 L or 21.5 mL.

So, to prepare 1.80 L of 0.215 M H2SO4, you would need to use 21.5 mL of 18 M sulfuric acid solution.

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Stab pull is the process by which the Earth's tectonic plates move apart at their edges due to the force of gravity at divergent plate boundaries. It leads to the formation of mountains and volcanic activity.

Stab pull refers to the process by which the Earth's tectonic plates move apart at their edges due to the force of gravity. This occurs at divergent plate boundaries, where the edges of the leaves become denser and sink downwards and away from each other. As a result, new crust is created from magma upwelling beneath the leaves, leading to the formation of mountains and volcanic activity.

For example, the East African Rift System is a prime location for stab pull. The African Plate is splitting apart along this boundary, causing the formation of the Great Rift Valley and the creation of a new crust.

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The intermolecular forces present between [tex]{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}[/tex] molecules are  

[tex]\boxed{{\text{hydrogen bonding and dispersion forces}}}[/tex]

Further Explanation:

Intermolecular forces:

The forces that exist between the molecules are known as intermolecular forces (IMF). IMF includes both attractive as well as repulsive forces. They are electrostatic in nature and determine the bulk properties of the substances like melting and boiling points. Molecules are held in any substance due to these forces.

The various types of intermolecular forces are as follows:

1. Hydrogen bonding:

It is an attractive force that exists between hydrogen and more electronegative elements like N, O, F. It can either be intermolecular or intramolecular. Intermolecular hydrogen bonding is the one that occurs between different molecules. For example, the bond between HF and [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex]  is an intermolecular hydrogen bond. Intramolecular hydrogen bonding occurs between various parts of the same molecule. Ortho-nitro phenol and salicylaldehyde show this type of bonding.

2. Ion-dipole forces:

It is an attractive force that occurs between an ion and a molecule consisting of a dipole. The force between [tex]{\text{N}}{{\text{a}}^ + }[/tex]  and water molecule is an example of this force.

3. Ion-induced dipole forces:

It is an attractive force that occurs between an ion and a nonpolar molecule. It induces a dipole in the molecule, resulting in ion-induced dipole force. The bond between [tex]{\text{F}}{{\text{e}}^{2 + }}[/tex]  and oxygen molecule is an example of such kind of bond.

In [tex]{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}{\mathbf{C}}{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{N}}{{\mathbf{H}}_{\mathbf{2}}}[/tex] , there is a presence of amine group [tex]\left({{\text{ - N}}{{\text{H}}_2}}\right)[/tex]  having a highly electronegative N atom. So it can form hydrogen bonds with hydrogen atom of solvent molecule like water. Also, a hydrocarbon chain [tex]\left({{\text{ - C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}}\right)[/tex]  is present in the molecule and we know dispersion forces exist between atoms and molecules. So [tex]{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}[/tex]  has both dispersion forces as well as hydrogen bonds between its molecules.

Learn more:

1. Number of covalent bonds formed by nitrogen: https://brainly.com/question/6029316

2. Which pair will form covalent bond?: https://brainly.com/question/1126757

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic and covalent bonding

Keywords: Intermolecular forces, CH3CHNH2, hydrogen bonding, dispersion forces, -NH2, -CH3CH2, amine, hydrocarbon, atoms, molecules.

The calculated mass of [tex]Al_2O_3[/tex] formed should be approximately 223.04 g.

Given reaction: [tex]2Al + Fe_2O_3[/tex] → [tex]Al_2O_3 + 2Fe[/tex]

Calculate the moles of Al and [tex]Fe_2O_3[/tex]:

Molar mass of Al = 26.98 g/mol

Molar mass of [tex]Fe_2O_3[/tex] = (55.85 g/mol * 2) + (16.00 g/mol * 3)

= 159.70 g/mol

Moles of Al = 118 g / 26.98 g/mol

≈ 4.37 moles

Moles of [tex]Fe_2O_3[/tex] = 601 g / 159.70 g/mol

≈ 3.76 moles

Determine the limiting reactant:

The balanced equation shows that 2 moles of Al react with 1 mole of [tex]Fe_2O_3[/tex]. So, for the given moles of [tex]Fe_2O_3[/tex], Al is the limiting reactant.

Calculate the moles of [tex]Al_2O_3[/tex] formed:

From the balanced equation, 2 moles of Al produce 1 mole of [tex]Al_2O_3[/tex].

Moles of [tex]Al_2O_3[/tex] formed = 4.37 moles Al / 2

≈ 2.19 moles

Calculate the mass of [tex]Al_2O_3[/tex] formed:

Molar mass of [tex]Al_2O_3[/tex] = (26.98 g/mol * 2) + (16.00 g/mol * 3)

= 101.96 g/mol

Mass of [tex]Al_2O_3[/tex] formed = 2.19 moles * 101.96 g/mol

≈ 223.04 g

Therefore, the calculated mass of [tex]Al_2O_3[/tex] formed should be approximately 223.04 g.

Learn more about limiting reagent, here:

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To calculate the mass of Al2O3 formed and the amount of excess reagent left in the reaction between aluminum and iron(III) oxide, we need to perform stoichiometric calculations that involve converting grams to moles, identifying the limiting reagent, applying stoichiometric ratios to determine product amounts, and converting moles back to grams.

To answer this question, we need to apply the concept of stoichiometry, which is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. As the balanced chemical equation for the reaction between aluminum and iron(III) oxide is given as 2Al + Fe2O3 → Al2O3 + 2Fe, this suggests that 2 moles of aluminum react with 1 mole of iron(III) oxide to produce 1 mole of aluminum oxide and 2 moles of iron.

To find the mass of Al2O3 formed, we would convert the given mass of Al and Fe2O3 to moles using their respective molar masses, see which reagent is the limiting reagent, and then use this information to determine the mass of Al2O3 formed in the reaction. The limiting reagent in a chemical reaction is the reactant that is completely consumed and determines the maximum amount of product that can be formed. Once the limiting reagent is consumed, the reaction stops, leaving the other reactant in excess.

By converting the remaining mass of the excess reagent left at the end of the reaction back to grams, we would calculate the mass of the excess reagent left. This involves a multi-step process encompassing the conversion of mass to moles, comparison of molar amounts, application of stoichiometry to find product amounts, and conversion of moles back to grams.

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Answer:

i think its C

Explanation: