Calculate the concentration of nitrate ion when dissolving 25.0 g of cobalt(II) nitrate Co(NO3)2 in 0.50 L aqueous solution. [MM CO(NO3)2 = 182.95 g/mol]

The volume of the gold bar pf dimension 14 cm×8 cm×4 cm is 448 cm³ and 448 mL or 0.448 L.

The volume of a rectangular prism is calculated by multiplying the length, width, and height. In this case, the length is 14 cm, the width is 8 cm, and the height is 4 cm. To calculate the volume of the gold bar, we use the formula V = l × w × h, where l, w, and h represent the length, width, and height of the bar, respectively. Plugging in the given dimensions, we have V = 14 cm × 8 cm × 4 cm = 448 cm³. Since 1 cm³ is equivalent to 1 mL, the volume of the gold bar is also 448 mL.

The volume of the gold bar, calculated using its given dimensions, is 448 cm³ and 448 mL. This volume represents the amount of space occupied by the gold bar.

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The balanced chemical equation of the Haber process is:

N2 + 3H2 → 2NH3

To calculate the mass of ammonia produced when 35.0g of nitrogen reacts with 12.5 g of hydrogen using the Haber process, we need to find the limiting reactant first.

Limiting reactant is the reactant which gets completely consumed in a chemical reaction, limiting the amount of product produced. Therefore, we must calculate the moles of each reactant using their molar masses and compare them to find the limiting reactant.

For nitrogen, the molar mass = 28 g/mol

Number of moles of nitrogen = 35.0 g / 28 g/mol = 1.25 mol

For hydrogen, the molar mass = 2 g/mol

Number of moles of hydrogen = 12.5 g / 2 g/mol = 6.25 mol

From the above calculations, it can be observed that hydrogen is in excess as it produces more moles of NH3. Thus, nitrogen is the limiting reactant.

Using the balanced chemical equation, the number of moles of NH3 produced can be calculated.

Number of moles of NH3 = (1.25 mol N2) × (2 mol NH3/1 mol N2) = 2.50 mol NH3Now,

to find the mass of NH3 produced, we can use its molar mass which is 17 g/mol.Mass of NH3 produced = (2.50 mol NH3) × (17 g/mol) = 42.5 g

Therefore, the mass of ammonia produced when 35.0g of nitrogen reacts with 12.5 g of hydrogen using the Haber process is 42.5 g.

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Yes, HCl is a strong acid and thus it does not have a conjugate base.

But, when HCl gets dissolved in water, it gives H+ and Cl- ions as its products. Here, Cl- acts as the conjugate base of HCl. Thus, HCl and Cl- form a conjugate acid-base pair. Therefore, the answer is: yes, HCl and Cl- form a conjugate acid-base pair.HCl is a strong acid and thus it does not have a conjugate base. But, when HCl gets dissolved in water, it gives H+ and Cl- ions as its products. Here, Cl- acts as the conjugate base of HCl. Thus, HCl and Cl- form a conjugate acid-base pair. Therefore, the answer is: yes, HCl and Cl- form a conjugate acid-base pair.

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A. the pressure exerted by 1.00 mol Cl2 confined to 3.00 L at 25°C is 8.12 atm. The answer to part a) is 8.12 atm.

B. the pressure exerted by 1.00 mol Cl2 confined to 3.00 L at 25°C is 8.12 atm. The answer to part a) is 8.12 atm.

C.  the difference between the result for an ideal gas and that calculated using the van der Waals equation is greater when the gas is confined to 3.00 L compared to 22.4 L.

a) Use the ideal gas equation:

The ideal gas equation is given by PV = nRT, where

P = pressure of gas

V = volume of gas

n = number of moles of gas

R = gas constant

T = temperature of gas

The pressure exerted by 1.00 mol Cl2 confined to a volume of 3.00 L at 25°C can be calculated using the ideal gas equation. The gas constant R in this equation is 0.0821 L atm/mol K (since volume is in liters and pressure is in atmospheres).

n = 1.00 mol

R = 0.0821 L atm/mol K

P = ?

V = 3.00 L (Volume)

T = 25 + 273 = 298 K (Temperature)

We can solve for P:

PV = nRT

P = (nRT) / V = (1.00 mol)(0.0821 L atm/mol K)(298 K) / (3.00 L)

P = 8.12 atm

Thus, the pressure exerted by 1.00 mol Cl2 confined to 3.00 L at 25°C is 8.12 atm. The answer to part a) is 8.12 atm.

b) Use van der Waals equation for your calculation:

The van der Waals equation is given by

(P + a(n/V)^2)(V - nb) = nRT

where a and b are van der Waals constants that depend on the gas. Values for the van der Waals constants are a = 6.49, b = 0.0562.

Using these values, we can calculate the pressure exerted by 1.00 mol Cl2 confined to a volume of 3.00 L at 25°C. The van der Waals constant R in this equation is 0.0821 L atm/mol K.

n = 1.00 mol

R = 0.0821 L atm/mol K

(P + a(n/V)^2) = nRT / (V - nb)

P = nRT / (V - nb) - a(n/V)^2

P = (1.00 mol)(0.0821 L atm/mol K)(298 K) / (3.00 L - (1.00 mol)(0.0562 L/mol)) - 6.49 atm (1.00 mol / (3.00 L)^2)

P = 7.73 atm

Thus, the pressure exerted by 1.00 mol Cl2 confined to a volume of 3.00 L at 25°C, as calculated using the van der Waals equation, is 7.73 atm. The answer to part b) is 7.73 atm.

c)The ideal gas law assumes that gas molecules have zero volume and do not interact with each other. The van der Waals equation accounts for non-ideal behavior by including the volume and attractive forces of gas molecules.

When a gas is confined to a small volume, the volume occupied by the gas molecules becomes more significant, and the attractive forces between molecules become stronger.

Thus, the difference between the result for an ideal gas and that calculated using the van der Waals equation is greater when the gas is confined to 3.00 L compared to 22.4 L.

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The current corresponding to the basal rate of oxygen consumption of a 70-kg person, which is 16 mol O2 per day, is approximately 0.19 Amperes.

To calculate the current, we need to convert the number of moles of oxygen consumed to the number of electrons involved in the reduction of oxygen.

From the balanced equation: O2 + 4H+ + 4e- → 2H2O, we can see that for every 4 moles of oxygen consumed, 4 moles of electrons are involved.

Therefore, the number of moles of electrons involved in the reduction of oxygen is also 16 mol.

To calculate the charge in coulombs (C), we use Faraday's constant (F) which is equal to 96485 C/mol.

Charge (C) = moles of electrons × Faraday's constant

Charge = 16 mol × 96485 C/mol

Charge ≈ 1543760 C

Finally, to calculate the current (I) in Amperes (A), we divide the charge by the time in seconds. Assuming a day consists of 24 hours (86400 seconds), we have:

Current (A) = Charge (C) / Time (s)

Current ≈ 1543760 C / 86400 s

Current ≈ 17.86 A

Therefore, the current corresponding to the basal rate of oxygen consumption of a 70-kg person is approximately 0.19 Amperes.

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The percent yield for each compound:

a) For NaHCO3: 100%
b) For Na2CO3: 126.2%
c) For CO2: 100%
d) For H2O: 0%

To find the percent yield, we first need to determine the theoretical yield and actual yield for each compound. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly, while the actual yield is the amount of product obtained in reality.

Let's calculate the theoretical yield for each compound:

a) NaHCO3(s): Since 3.08 g of NaHCO3 is given, the theoretical yield of NaHCO3 would also be 3.08 g.

b) Na2CO3(s): The given problem states that 1.04 g of Na2CO3 is obtained. However, since Na2CO3 is formed from NaHCO3, we need to consider the molar mass ratio between NaHCO3 and Na2CO3. The molar mass of NaHCO3 is 84 g/mol, and the molar mass of Na2CO3 is 106 g/mol. Using this ratio, we can calculate the theoretical yield of Na2CO3:

(1.04 g Na2CO3) × (84 g NaHCO3 / 106 g Na2CO3) = 0.824 g NaHCO3

c) CO2(g): CO2 is produced during the decomposition of NaHCO3, and it is a gas. Therefore, we need to convert the mass of NaHCO3 to moles and then use the balanced chemical equation to find the moles of CO2 produced. The balanced equation for the decomposition of NaHCO3 is:

2 NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)
The molar mass of NaHCO3 is 84 g/mol.

(3.08 g NaHCO3) / (84 g/mol NaHCO3) = 0.0367 mol NaHCO3
According to the balanced equation, 1 mole of NaHCO3 produces 1 mole of CO2. Therefore, the theoretical yield of CO2 is also 0.0367 mol.

d) H2O(g): Similarly, we can use the balanced equation to determine the theoretical yield of water. According to the equation, 1 mole of NaHCO3 produces 1 mole of H2O. Therefore, the theoretical yield of H2O is 0.0367 mol.

Now, let's calculate the percent yield for each compound:

Percent yield = (Actual yield / Theoretical yield) × 100

a) For NaHCO3:
Percent yield = (3.08 g / 3.08 g) × 100 = 100%

b) For Na2CO3:
Percent yield = (1.04 g / 0.824 g) × 100 = 126.2%

c) For CO2:
Percent yield = (0.0367 mol / 0.0367 mol) × 100 = 100%

d) For H2O:
Percent yield = (0 mol / 0.0367 mol) × 100 = 0%

To summarize, the percent yield for NaHCO3 is 100%, for Na2CO3 is 126.2%, for CO2 is 100%, and for H2O is 0%.

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