What are salt ? How they are used

ang mga kumpanya ng electric motorcycle ay nag anunsyon na hindi na sila gagawa pa ng motor na pinaandar ng kuryente .ano sa pal

andreev551 [17]

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5 0

3 years ago

What is the percent composition of dinotrogen pentoxide? With work shown please

iris [78.8K]

Answer:

%N = 25.94%

%O = 74.06%

Explanation:

Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅

We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.

m(N): 2 × 14.01 g = 28.02 g

Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅

We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.

m(O): 5 × 16.00 g = 80.00 g

Step 3: Calculate the mass of 1 mole of N₂O₅

We will sum the masses of N and O.

m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g

Step 4: Calculate the percent composition of N₂O₅

We will use the following expression.

%Element = m(Element)/m(Compound) × 100%

%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%

%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%

7 0

2 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

PolarNik [594]

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

6 0

3 years ago

The element least likely to form a bond is group

guapka [62]

Between phosphorous, sulfur, chlorine and argon, argon is the element that's the least likely to form an ionic bond with sodium.

7 0

3 years ago

How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this

KiRa [710]

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = $\frac{mass}{molar mass}$

Number of moles of acetone = $\frac{31.5}{58.08}$

Number of moles of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

4 0

3 years ago