To make a 1.75 M solution of Ba(NO₃)₂ with a volume of 2.50 L, you will need 1141.72 grams of the solute.
Firstly, we need to understand that Molarity (M) is defined as the number of moles of solute per liter of solution. Thus, we can use the formula:
Molarity (M) = (Number of moles of solute) / (Volume of solution in liters)
We have been given the volume of the solution (V) as 2.50 L and the Molarity (M) as 1.75 M. We need to find out the number of moles of solute (n) required to prepare this solution.
Rearranging the above formula, we get:
Number of moles of solute = Molarity × Volume of solution in liters
Substituting the given values, we get:
Number of moles of solute = 1.75 mol/L × 2.50 L = 4.375 mol
The molecular weight of Ba(NO₃)₂ can be calculated by adding the atomic weights of its constituents, which are Ba=137.33 g/mol, N=14.01 g/mol, O=16.00 g/mol. Thus, the molecular weight of Ba(NO₃)₂ comes out to be:
Molecular weight of Ba(NO₃)₂ = (137.33 g/mol) + 2 × (14.01 g/mol + 3 × 16.00 g/mol) = 261.34 g/mol
Now we can use the formula:
Mass of solute (in grams) = Number of moles of solute × Molecular weight of solute
Substituting the values, we get:
Mass of solute (in grams) = 4.375 mol × 261.34 g/mol = 1141.72 g
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. ethanol (ch3ch2oh) burns in air to generate carbon dioxide and water, a. write a balanced equation to show this reaction b. determine the volume of air (not oxygen) in liters at 35 degrees c and 790 mm hg required to burn 250 grams of ethanol.
(a). [tex]C_2H_5OH + 3O_2[/tex] → [tex]2CO_2 + 3H_2O[/tex]
(b). The volume of air required to burn 250 grams of ethanol at 35°C and 790 mmHg is approximately 6.63 liters.
a. The balanced equation for the combustion of ethanol ([tex]C_2H_5OH[/tex]) in air to generate carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]) is:
[tex]C_2H_5OH + 3O_2[/tex] → [tex]2CO_2 + 3H_2O[/tex]
b. We first need to calculate the number of moles of ethanol used in the reaction. The molar mass of ethanol is:
46.07 g/mol
Therefore, the number of moles of ethanol used is:
[tex]n = m/M = 250 g / 46.07 g/mol = 5.42 mol[/tex]
Therefore, the number of moles of oxygen required to burn 5.42 moles of ethanol is:
[tex]3n = 3 * 5.42 mol = 16.26 mol[/tex]
The ideal gas law is:
PV = nRT
V = nRT/P
Substituting the values, we get:
[tex]V = (16.26 mol)(0.08206 L.atm/(mol.K))(308.15 K) / 790 mmHg[/tex]
Simplifying, we get:
V = 6.63 L
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Find the molarity of 194. 55 g of sugar (C12H22O11) in 250. ML of water
To find the molarity of a solution, we need to know the number of moles of solute (in this case, sugar) and the volume of the solution in liters. We can use the given mass of sugar and the molar mass of sugar to find the number of moles:
Mass of sugar = 194.55 g
Molar mass of sugar (C12H22O11) = 342.3 g/mol
Number of moles of sugar = Mass of sugar / Molar mass of sugar
Number of moles of sugar = 194.55 g / 342.3 g/mol
Number of moles of sugar = 0.568 mol
Now we need to convert the given volume of the solution (250 mL) to liters:
Volume of solution = 250 mL
Volume of solution = 250 mL / 1000 mL/L
Volume of solution = 0.250 L
Finally, we can use the number of moles of sugar and the volume of the solution to calculate the molarity:
Molarity = Number of moles of sugar / Volume of solution
Molarity = 0.568 mol / 0.250 L
Molarity = 2.27 M
Therefore, the molarity of the sugar solution is 2.27 M.
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Please Help!!!! D:
A student runs tests on an unknown substance and discovers the following properties. What other property does this element most likely have?
A highly reactive
B low electronegativity
C has many isotopes
D not found pure in nature
The unknown substance most likely has property not found pure in nature.(D)
Since the substance has properties A (highly reactive) and B (low electronegativity), it's likely that it readily forms compounds with other elements, making it difficult to find in its pure form.
Highly reactive elements, such as alkali metals or halogens, are typically not found in nature in their pure state because they readily react with other elements to form stable compounds. Property C (has many isotopes) doesn't directly influence the substance's reactivity or occurrence in nature.(D)
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⁻⁻⁻⁻⁻ results in a new substance and it cannot be reversed by physical means
Chemical change results in a new substance and it cannot be reversed by physical means.
What is chemical change?Chemical changes is said to occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances and are not reversible except by further chemical reactions.
Examples of chemical change would be:
Burning a piece of paper would be a chemical change, and also baking a cake.It is also worthy to note that in a physical change, no new substance is formed and also a chemical change is always accompanied by one or more new substance.
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How many grams of oxygen (O2) is required to burn 28. 8 g of ammonia (NH3)?
4NH3 + 7O2 → 4NO2 + 6H2O
Molar Masses
NH3=17. 0305 g/mol
O2=31. 998 g/mol
NO2=46. 0055 g/mol
H2O=18. 0153 g/mol
a)15. 3 g
b)94. 9 g
c)54. 1 g
d)108 g
The number of grams of oxygen required is 94.9 g, under the condition that it is used to burn 28. 8 g of ammonia (NH₃)
NH₃ + 7O₂ → 4NO₂ + 6H₂O,
then the correct answer for the required question is Option B.
Now, the balanced chemical equation for the reaction of ammonia (NH₃) and oxygen (O₂) to create nitrogen dioxide (NO₂) and water (H₂O) is
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
The given molar mass of NH₃ is 17.0305 g/mol and that of O₂ is 31.998 g/mol.
In order to find out how many grams of O₂ are required to burn 28.8 g of NH₃, we have to first balance the equation:
4 NH₃+ 7O₂ → 4NO₂ + 6H₂O
Then there are 4 moles of NH₃, we need 7 moles of O₂.
Hence, molar mass of NH₃ is 17.0305 g/mol, so we can change 28.8 g of NH₃ to moles
28.8 g NH₃ × (1 mol NH₃/17.0305 g NH₃)
= 1.69 mol NH₃
Now we have to apply stoichiometry to evaluate how many moles of O₂ are required
1.69 mol NH₃ × (7 mol O₂/4 mol NH₃)
= 2.95 mol O₂
Therefore, we can convert moles of O₂ to grams:
2.95 mol O₂ × (31.998 g O₂/1 mol O₂)
= 94.9 g
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The complete question is
How many grams of oxygen (O2) is required to burn 28. 8 g of ammonia (NH3)?4NH3 + 7O2 → 4NO2 + 6H2O
Molar Mass
NH3=17. 0305 g/mol
O2=31. 998 g/mol
NO2=46. 0055 g/mol
H2O=18. 0153 g/mol
a)15. 3 g
b)94. 9 g
c)54. 1 g
d)108 g
the total volume of hydrogen gas needed to fill the hindenburg was l at atm and . given that for is , how much heat was evolved when the hindenburg exploded, assuming all of the hydrogen reacted to form water?
2.4453 × 10⁹ KJ energy was evolved when the total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 10⁸ l at 1.00 atm and 25.1°
According to the given data,
Volume of the hydrogen gas = 2.09 × 10⁸ L
Pressure of the gas = P = 1 atm
Temperature of the gas =T = 25.1 °C =298.1 K
We know that, for an ideal gas equation
PV=nRT
1 atm ×2.09 × 10⁸ L = n × 0.0820 atmL/molK × 298.1 K
⇒n = 1 atm ×2.09 × 10⁸ L/ 0.0820 atmL/molK × 298.1 K
⇒n = 0.0855 ×10⁸ mol
ΔH for hydrogen gas is =-286 kJ/mol
For 0.0855 ×10⁸ mol energy evolved when hydrogen gas is burned =
0.0855 ×10⁸ mol × (-286 KJ/mol) = -2.4453 × 10⁹ KJ
Therefore, 2.4453 × 10⁹ KJ energy was evolved when it was burned.
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The complete question is-
The total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 108 l at 1.00 atm and 25.1°. how much energy was evolved when it burned?
A 3.950 l sample of gas is cooled from 91.50°c to a temperature at which its volume is 2.550 l. what is this new temperature? assume no change in pressure of the gas.
When a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L with no change in pressure, the new temperature is approximately -36.61°C.
To find the new temperature when a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L, we can use the Charles' Law formula. Charles' Law states that the volume of a gas is directly proportional to its temperature, assuming that pressure remains constant.
Mathematically, this can be represented as:
V1/T1 = V2/T2
Here, V1 is the initial volume (3.950 L), T1 is the initial temperature (91.50°C), V2 is the final volume (2.550 L), and T2 is the final temperature, which we need to find.
First, convert the initial temperature from Celsius to Kelvin by adding 273.15:
T1 = 91.50°C + 273.15 = 364.65 K
Now, plug the values into the Charles' Law formula:
(3.950 L) / (364.65 K) = (2.550 L) / T2
To find T2, we can cross-multiply and divide:
T2 = (2.550 L) * (364.65 K) / (3.950 L)
T2 ≈ 236.54 K
Finally, convert the temperature back to Celsius by subtracting 273.15:
New temperature = 236.54 K - 273.15 ≈ -36.61°C
In conclusion, when a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L with no change in pressure, the new temperature is approximately -36.61°C.
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How many compounds are there in 434g of ammonium nitrate?
3.266 × 10²⁴ compounds in 434g of ammonium nitrate
To determine how many compounds are in 434g of ammonium nitrate, we will follow these steps:
Step 1: Determine the molar mass of ammonium nitrate (NH₄NO₃).
Ammonium nitrate consists of one nitrogen (N) atom, four hydrogen (H) atoms, and three oxygen (O) atoms in its chemical formula. The molar masses of N, H, and O are approximately 14 g/mol, 1 g/mol, and 16 g/mol, respectively.
Molar mass of NH₄NO₃ = 1(N) + 4(H) + 1(N) + 3(O)
= 14 + (4 × 1) + 14 + (3 × 16)
= 14 + 4 + 14 + 48
= 80 g/mol
Step 2: Calculate the number of moles of ammonium nitrate.
To find the number of moles, divide the given mass (434g) by the molar mass (80 g/mol).
Number of moles = 434 g / 80 g/mol
= 5.425 moles
Step 3: Calculate the number of compounds (molecules) in ammonium nitrate.
Use Avogadro's number (6.022 × 10²³ molecules/mol) to find the total number of molecules in 5.425 moles of ammonium nitrate.
Number of compounds = 5.425 moles × (6.022 × 10²³ molecules/mol)
= 3.266 × 10²⁴ molecules
So, there are approximately 3.266 × 10²⁴ compounds in 434g of ammonium nitrate.
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