I would say that this problem needs more information. Giving only the concentration would not specify anything since in general salts are neutral as they are from neutralization reactions. Specifying other properties of the compound would be more helpful.
Mass of carbon = 0.1927 g
Mass of hydrogen = 0.02590 g
Mass of nitrogen = 0.1124 g
Mass of phosphorus = 0.1491 g
Mass of Oxygen = 0.8138 - ( 0.1927 + 0.02590 + 0.1124 + 0.1491 ) = 0.3337 g
Moles of C = ![\frac{0.1927}{12.011} = 0.0160 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.1927%7D%7B12.011%7D%20%3D%200.0160%20mol)
Moles of H =
mol
Moles of O = ![\frac{0.3337}{15.999} = 0.0209 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.3337%7D%7B15.999%7D%20%20%3D%200.0209%20mol)
Moles of P = ![\frac{0.1491}{30.974} = 0.00481 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.1491%7D%7B30.974%7D%20%3D%200.00481%20mol)
Moles of N = ![\frac{0.1124}{14.01} = 0.00802 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.1124%7D%7B14.01%7D%20%20%3D%200.00802%20mol)
So, the empirical formula will be: ![C_{0.0160} H_{0.0257} O_{0.0209} N_{0.00802} P_{0.00481}](https://tex.z-dn.net/?f=C_%7B0.0160%7D%20H_%7B0.0257%7D%20O_%7B0.0209%7D%20N_%7B0.00802%7D%20P_%7B0.00481%7D)
To make it a whole divide the moles of all elements by 0.00481.
Therefore,
.
Molecular mass of ATP = 507 g/mol
Empirical formula = 164.063 g/mol
Molecular formula:
![\frac{507}{164} = 3](https://tex.z-dn.net/?f=%5Cfrac%7B507%7D%7B164%7D%20%3D%203)
So, molecular formula is
= ![C_1_0 H_1_6 O_1_3 N_5P_3](https://tex.z-dn.net/?f=C_1_0%20H_1_6%20O_1_3%20N_5P_3)
To learn more about molecular formula visit: brainly.com/question/14425626
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When you are burning a log <span>carbon dioxide and water are formed.</span>
<span>HCl+NaOH---> NaCl+H2O
the 2 reactives are monovalent . the NaOH is 3 times as concentrated as HCl
So you need 3 volumes HCl 2M to neutralize 1 volume NaOH 6M
and for 75 ml NaOH 6M you need 75*3= 225mL HCl 2M
second question HBr+NH4OH--> NH4Br+H2O
as before 1 HBr for 1NH4OH is needed IF they have the same molarity
Here you need 40/10 =4 more HBr than NaOH
The molarity of NaOH is 4*0.5 =2M</span>
Answer:
1.2 M
1.2 m
Explanation:
There is some info missing. I think this is the original question.
<em>A student dissolves 5.1g of ammonia (NH₃) in 250 mL of a solvent with a density of 1.02 g/mL. The student notices that the volume of the solvent does not change when the ammonia dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>
<em />
Molarity
The molar mass of ammonia is 17.03 g/mol. The moles corresponding to 5.1 grams is:
5.1 g × (1 mol/17.03 g) = 0.30 mol
The volume of the solution is equal to the volume of the solvent, 250 mL or 0,250 L.
The molarity of ammonia is:
M = moles of solute / liters of solution
M = 0.30 mol / 0.250 L
M = 1.2 M
Molality
The density of the solvent is 1.02 g/mL. The mass corresponding to 250 mL is:
250 mL × 1.02 g/mL = 255 g = 0.255 kg
The molality of ammonia is:
m = moles of solute / kilograms of solvent
m = 0.30 mol / 0.255 kg
m = 1.17 m ≈ 1.2 m