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What is keq for the reaction 2HCI(g) H2(g) plus Cl2(g)?
For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
How many milliliters of Cl2 gas must you have to obtain 0.30g at STP in chemistry?
You can use the ideal gas law to solve this problem. First, convert 0.30 g of Cl2 to moles. Then use the molar volume of gas at STP (22.4 L/mol) to determine the volume of Cl2 gas needed. Convert this volume to milliliters (1 L = 1000 mL) to find the answer.
What is the volume of an object if the density is 0.45 g ml and the mass is 25 g?
The volume of the object can be calculated using the formula: volume = mass / density. Plugging in the values (mass = 25 g, density = 0.45 g/ml), the volume would be 55.56 ml.
What is the density of an object with a mass of 500 g and has a volume of 25 cm?
The density of the object is 20 g/cm^3. Density is calculated by dividing the mass of the object by its volume. In this case, 500g ÷ 25cm^3 = 20 g/cm^3.
How Many Moles Of PCl5 Can Be Produced From 58.0 G Of Cl2 (and Excess P4)?
To determine how many moles of PCl5 can be produced from 58.0 g of Cl2, we first need to calculate the moles of Cl2. The molar mass of Cl2 is approximately 70.9 g/mol, so the number of moles of Cl2 is 58.0 g / 70.9 g/mol ≈ 0.819 moles. The balanced reaction for the formation of PCl5 from P4 and Cl2 is: P4 + 10 Cl2 → 4 PCl5. From this, we see that 10 moles of Cl2 produce 4 moles of PCl5, so 0.819 moles of Cl2 can produce (0.819 moles Cl2) * (4 moles PCl5 / 10 moles Cl2) ≈ 0.3276 moles of PCl5. Thus, approximately 0.328 moles of PCl5 can be produced.
What is keq for the reaction 2HCI(g) H2(g) plus Cl2(g)?
For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
How many milliliters of Cl2 gas must you have to obtain 0.30g at STP in chemistry?
You can use the ideal gas law to solve this problem. First, convert 0.30 g of Cl2 to moles. Then use the molar volume of gas at STP (22.4 L/mol) to determine the volume of Cl2 gas needed. Convert this volume to milliliters (1 L = 1000 mL) to find the answer.
What is Keq for the reaction H2(g) plus Cl2(g) 2HCl(g)?
For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
What is he necessary amount of chlorine needed to prepare 25.00 g of hydrogen chloride by reacting hydrogen with chlorine?
H2 (g) + Cl2 (g) --> 2 HCl (g) 25.00 g HCl x 1 mol HCl x 1 mol Cl2 x 70.90 g Cl2 = 24.3 g Cl2 are needed. ................... 36.46 g HCl . 2 mol HCl .. 1 mol Cl2
Write a balanced equation for the reaction of chlorine gas with fluorine gas?
Cl2(g) + F2(g) -> 2ClF(g)
What is the volume of an object if the density is 0.45 g ml and the mass is 25 g?
The volume of the object can be calculated using the formula: volume = mass / density. Plugging in the values (mass = 25 g, density = 0.45 g/ml), the volume would be 55.56 ml.
What is the density for a 25 ml of water with a mass of 25 g?
The formula for density is density = mass/volume. In this case, the mass is 25 g and the volume is 25 ml. Therefore, the density of water in this scenario would be 1 g/ml.
What is k eq for the reaction 2HCI(g) H2(g) plus CI2(g)?
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
A piece of lead has a density of 11.3 g per cm cubed and a mass of 282.5 g. What is its volume?
To find the volume, you can use the formula: Volume = Mass / Density. Plugging in the values gives Volume = 282.5 g / 11.3 g/cm^3 = 25 cm^3. So, the volume of the lead is 25 cm^3.
What is the density of an object with a mass of 500 g and has a volume of 25 cm?
The density of the object is 20 g/cm^3. Density is calculated by dividing the mass of the object by its volume. In this case, 500g ÷ 25cm^3 = 20 g/cm^3.
What is the density of a substance with a mass of 25 g and a volume of 100cm3?
The density of the substance is 0.25 g/cm^3. This is calculated by dividing the mass (25 g) by the volume (100 cm^3).
How Many Moles Of PCl5 Can Be Produced From 58.0 G Of Cl2 (and Excess P4)?
To determine how many moles of PCl5 can be produced from 58.0 g of Cl2, we first need to calculate the moles of Cl2. The molar mass of Cl2 is approximately 70.9 g/mol, so the number of moles of Cl2 is 58.0 g / 70.9 g/mol ≈ 0.819 moles. The balanced reaction for the formation of PCl5 from P4 and Cl2 is: P4 + 10 Cl2 → 4 PCl5. From this, we see that 10 moles of Cl2 produce 4 moles of PCl5, so 0.819 moles of Cl2 can produce (0.819 moles Cl2) * (4 moles PCl5 / 10 moles Cl2) ≈ 0.3276 moles of PCl5. Thus, approximately 0.328 moles of PCl5 can be produced.
