If a sample of hydrochloric acid is neutralized with sodium hydroxide, what products are formed?

Indica, con base en la siguiente ecuación:

Anvisha [2.4K]

Hello I just want to say have a good day

7 0

3 years ago

How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102

Ede4ka [16]

Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles$

1 mole of hydrogen $(H_2)$ = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms

17.5 mole of hydrogen $(H_2)$ = $\frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25}$ atoms

There are $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0

3 years ago

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

Total pressure = partial pressure of nitrogen + partial pressure of oxygen

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: P₁/T₁=P₂/T₂

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0

3 years ago

1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a

Archy [21]

Answer:

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃ + Al → 3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

Al : Ag

1 : 3

0.223 : 3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂ → 2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

HgO : Hg

2 : 2

3.12 : 3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂ + 5O₂ → 2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

N₂O₅ : O₂

2 : 5

12.99 : 5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

C₆H₆ : CO₂

2 : 12

0.103 : 12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0

3 years ago

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

Oxana [17]

12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g

7 0

4 years ago