Which of the following is an example of an intensive physical property?

Compound 1 has a composition of 46.7 mass % of element A and 53.3 mass % of element B. A and B also form a second binary compoun

Leviafan [203]

A summary of the Law of multiple proportions is that if A and B form more than one compound, and B1 is the amount of element B which reacts with a fixed mass of A in compound 1, and B2 is the amount of B which reacts with the same fixed mass of B to form compound 2, then the ratio B1:B2 will be small whole numbers.

This law is rather simplistic, and given the range of compounds known today the definition of 'small' is now rather large... but, to answer the question:

in compound one 1.14133g of B reacts with 1g of A. (1.14133=53.3/46.7)

5 0

3 years ago

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

4 years ago

How many moles of sodium chloride can react with 18.3 liters of fluorine gas at 1.2 atmospheres and 299 Kelvin?

My name is Ann [436]

Answer:

1.79 mol.

Explanation:

For the balanced reaction:

2NaCl + F₂ → 2NaF + Cl₂.

It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.

Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = 18.3 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (299 K).

∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.

Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:

Using cross multiplication:

2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.

??? mol of NaCl is needed to react with → 0.895 mol of F₂.

∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.

4 0

3 years ago

How many mols of a 3M NaOH solution are in 1L of the solution?

Alex

To calculate this, we need the Molarity formula. This formula tell us that Molarity, which is a concentration unit, is equal to the number of moles divided by the volume. In this question we already have the Molarity and the Volume, so let's build our equation:

C = n/V (You can see Molarity with the letter "C" because it means concentration)

3 = n/1

n = 1 * 3

n = 3 moles of NaOH

6 0

2 years ago

Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi

polet [3.4K]

Answer:

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO₂ + 2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

Copper Metal = Cu

Nitric Acid = HNO₃

Copper (II) Nitrate = Cu(NO₃)₂

Nitrogen Dioxide = NO₂

Water = H₂O

Step 2: Write down the unbalance Chemical equation,

Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

Cu + 3 HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

Cu + 3 HNO₃ → Cu(NO₃)₂ + NO₂ + 2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

Cu + 4 HNO₃ → Cu(NO₃)₂ + NO₂ + 2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO₂ + 2 H₂O

7 0

3 years ago