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The process that occurs when physical forces break rock into smaller pieces without changing the rock's chemical composition is

sp2606 [1]

Answer:

A) Mechanical Weathering

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Explanation:

Mechanical weathering breaks rocks into smaller pieces without changing their composition. Ice wedging and abrasion are two important processes of mechanical weathering. Chemical weathering breaks down rocks by forming new minerals that are stable at the Earth's surface

4 0

3 years ago

Which of the following pairs of elements is most likely to form an ionic bond? A. nitrogen and oxygen B. aluminum and nitrogen C

krok68 [10]

Answer:

B

Explanation:

When nitrogen and aluminum form an ionic bond, the formula of the ionic compound is AlN. Aluminum forms a cation Al⁺³ and nitrogen forms an anion N⁻³. When each cation of Al⁺³ combines with an anion of N⁻³, AlN or aluminum nitride is formed

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6 0

3 years ago

Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (

Nesterboy [21]

Balanced chemical equation for the reaction is:

2S $O_{2}$ (g) + $O_{2}$ (g)+ 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

Moles of $H_{2} S_{} O_{4}$ formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of S $O_{2}$ = 11.5 moles

moles of $H_{2} S_{} O_{4}$ = ?

Moles of $O_{2}$ needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2S $O_{2}$ (g) + $O_{2}$ (g) + 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of S $O_{2}$ , x moles of $H_{2} S_{} O_{4}$ is formed

$\frac{1}{2} =\frac{x}{11.5}$

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

7 0

3 years ago

1/2-life of pyruvic acid in the presence of aminotransferase enzyme (which converts it to alanine) was found to be 221 s. how lo

Ugo [173]

Answer:

After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.

Explanation:

The first order kinetics reaction is:

ln [A] = ln [A]₀ - kt

<em>Where [A] is concentration after t time, [A]₀ is intial concentration and k is reaction constant.</em>

To convert half-life to k you must use:

t(1/2) = ln 2 / K

221s = ln 2 / K

K = ln 2 / 221s

If [A] = 1/64, [A]₀ = 1:

ln [A] = ln [A]₀ - kt

ln (1/64) = ln 1 - 3.1364x10⁻³t

4.1588 = 3.1364x10⁻³s⁻¹t

1326s = t

<h3>After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.</h3>

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6 0

3 years ago

C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w

Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃ ΔH = –183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂ ΔH = 183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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