A metal x with relative atomic mass 56 forms an oxide with formula x2O3 . How many grams of metal will combine with 10g of oxygen?(o= 16)
?X + ?O2 --- > ? X2O3 well we will need at least 6 O atoms
4 X + 3 O2 ---> 2 X2 O3
so 10 g O
mols O * 16 grams/ mol = 10 grams
that is 10/16 = 0.625 mols O
every mol of X2O3 contains 3 mols of O
so we have 0.625/3 = 0.298 mols of X2O3 or 0.596 mols of X
.596 * 56 = ?
Good
Good, but I don't really understand
thanks
I need some explanation to decode this please
I am explanation to decode this pls
To calculate the concentration of Cl ions in a 0.3 M sodium chloride (NaCl) solution, we need to use the formula:
concentration (in g/dm3) = molarity (in mol/dm3) x formula weight (in g/mol)
The formula weight of NaCl is 58.44 g/mol, which means each mole of NaCl contains 1 mole of Cl ions. Therefore, the 0.3 M NaCl solution contains 0.3 moles of Cl ions per liter.
Now we can calculate the concentration of Cl ions in g/dm3:
concentration (in g/dm3) = 0.3 mol/dm3 x 35.45 g/mol
concentration = 10.635 g/dm3
Therefore, the concentration of Cl ions in a 0.3 M NaCl solution is 10.635 g/dm3.
35.8g of a mixture of potassium chloride and potassium trioxochlorate (v)were heated to a constant to a constant mass.if the residue weighed 24.9g what was the percentage mass of potassium chloride in the mixture.kcl is not decomposed onmheating
not clear enough
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23.3
Calculate the concentrations in g/dm3of CL ions in 0.3m sodium chloride solutions
Calculate the concentrations in g/dm3 of the No3ions in 0.75 trioxonitrate (v)acid
To calculate the concentration of nitrate (NO3-) ions in a 0.75 M trioxonitrate (V) acid (HNO3) solution, we need to first understand the chemical formula of trioxonitrate (V) acid.
Trioxonitrate(V) acid is also known as nitric acid (HNO3). In this acid, the hydrogen ion (H+) and the nitrate ion (NO3-) are in a 1:1 ratio. Therefore, the concentration of NO3- ions in a HNO3 solution is the same as the concentration of H+ ions.
The molar mass of HNO3 is 63.01 g/mol. To calculate the concentration of H+ ions, we can use the formula:
concentration (in mol/dm3) = molarity (in mol/dm3) x acid dissociation constant (Ka)
The Ka of HNO3 is approximately 25 x 10^-6.
So, the concentration of H+ ions in a 0.75 M HNO3 solution is:
concentration H+ = 0.75 mol/dm3 x 25 x 10^-6
concentration H+ = 1.875 x 10^-3 mol/dm3
Since the concentration of NO3- ions is the same as the concentration of H+ ions in this acid, the concentration of NO3- ions in a 0.75 M HNO3 solution is also:
concentration NO3- = 1.875 x 10^-3 mol/dm3
Now, we can calculate the concentration of NO3- ions in g/dm3:
concentration NO3- (in g/dm3) = concentration NO3- (in mol/dm3) x formula weight NO3-
The formula weight of NO3- is 62.00 g/mol.
concentration NO3- (in g/dm3) = 1.875 x 10^-3 mol/dm3 x 62.00 g/mol
concentration NO3- = 0.11625 g/dm3
Therefore, the concentration of NO3- ions in a 0.75 M HNO3 solution is 0.11625 g/dm3.
Calculate the concentrations in g/dm3 of so4ions in 3m potassium tetraoxosulphate(iv)solution
To calculate the concentration of sulfate (SO42-) ions in a 3 M potassium tetraoxosulfate(IV) (K2SO4) solution, we need to first understand the chemical formula of potassium tetraoxosulfate(IV).
Potassium tetraoxosulfate(IV) is also known as potassium sulfate (K2SO4). In this salt, the potassium ion (K+) and the sulfate ion (SO42-) are in a 2:1 ratio. Therefore, the concentration of SO42- ions in a K2SO4 solution is half of the concentration of K2SO4.
The molar mass of K2SO4 is 174.24 g/mol. To calculate the concentration of K2SO4, we can use the formula:
concentration (in mol/dm3) = molarity (in mol/dm3)
So, the concentration of K2SO4 in a 3 M solution is:
concentration K2SO4 = 3 mol/dm3
The concentration of SO42- ions is half of the concentration of K2SO4:
concentration SO42- = 1.5 M
Now, we can calculate the concentration of SO42- ions in g/dm3:
concentration SO42- (in g/dm3) = concentration SO42- (in mol/dm3) x formula weight SO42-
The formula weight of SO42- is 96.06 g/mol.
concentration SO42- (in g/dm3) = 1.5 mol/dm3 x 96.06 g/mol
concentration SO42- = 144.09 g/dm3
Therefore, the concentration of SO42- ions in a 3 M K2SO4 solution is 144.09 g/dm3.
We know that the weight of the residue (after heating the mixture) was 24.9 g. This residue contains all the potassium chloride and potassium trioxochlorate that was in the original mixture.
Let's assume that the weight of potassium chloride in the original mixture was x g.
Then, the weight of potassium trioxochlorate in the original mixture is (35.8 - x) g.
When we heat the mixture, the potassium trioxochlorate decomposes into potassium chloride and oxygen. However, we know that the KCl is not decomposed on heating.
So, the residue after heating consists of only KCl. Therefore:
x g KCl = 24.9 g
x = 24.9 g
This means that the original mixture contained 24.9 g of KCl and (35.8 - 24.9) = 10.9 g of KClO3.
Now we can calculate the percentage mass of KCl in the mixture:
% mass of KCl = (mass of KCl / total mass of mixture) x 100
% mass of KCl = (24.9 g / 35.8 g) x 100
% mass of KCl = 69.55%
Therefore, the percentage of KCl in the mixture is 69.55%.
An excess of divalent metal M was dissolved in a limited volume of hydrochloride acid .if 576cm of hydrogen were liberated at STP what was the mass of metal that produced this volume of hydrogen (m=24,H=1,molar volume of a gas at STP=22.4dm3
The balanced chemical equation for the reaction between a divalent metal M and hydrochloric acid (HCl) is:
M + 2HCl → MCl2 + H2
From the equation, we can see that 1 mole of metal produces 1 mole of hydrogen gas. Therefore, the number of moles of metal that produced 576 cm3 (0.576 dm3) of hydrogen gas at STP (standard temperature and pressure) can be calculated using the molar volume of gas at STP:
number of moles of M = volume of H2 gas / molar volume of gas at STP
number of moles of M = 0.576 dm3 / 22.4 dm3/mol
number of moles of M = 0.0257 mol
Since the metal was in excess, all the hydrochloric acid was consumed in the reaction, meaning that the number of moles of hydrochloric acid (HCl) is twice the number of moles of metal (M):
number of moles of HCl = 2 x number of moles of M
Now we can use the number of moles of HCl and the molar mass of HCl to calculate the mass of hydrochloric acid used:
mass of HCl = number of moles of HCl x molar mass of HCl
The molar mass of HCl is 36.5 g/mol (hydrogen has a molar mass of 1 g/mol, and chlorine has a molar mass of 35.5 g/mol).
number of moles of HCl = 2 x 0.0257 mol
number of moles of HCl = 0.0514 mol
mass of HCl = 0.0514 mol x 36.5 g/mol
mass of HCl = 1.87 g
Therefore, the mass of metal M that produced 576 cm3 of hydrogen gas at STP is equivalent to the moles of hydrogen produced, which is 0.0257 mol. We can use the molar mass of metal M to calculate its mass:
mass of M = number of moles of M x molar mass of M
We do not know the identity of the metal, so we cannot use its molar mass directly. However, we do know that it is a divalent metal (i.e. it has a valency of 2), which means that its molar mass is twice the molar mass of its oxide. Since oxides of metals usually have a known formula weight, we can use this information to calculate the molar mass of M:
Let x be the molar mass of the unknown metal, and let y be the molar mass of its oxide.
Then, we have:
2x = y
The balanced chemical equation for the reaction of the metal with oxygen to form its oxide is:
M + O2 → MO
From the equation, we can deduce that:
1 mole of metal M reacts with 1 mole of oxygen to form 1 mole of its oxide MO.
Therefore, the number of moles of the oxide MO that corresponds to the 0.0257 mol of metal M is also 0.0257 mol.
Using the formula weight of MO (which is not given in the question), we can calculate y and then use it to find x.
Once we know x, we can use the first equation above to calculate the molar mass of the metal M, and then use this molar mass to calculate its mass:
mass of M = number of moles of M x molar mass of M
I hope that helps!
What volume of 0.05mol/dm3 NaCl
I'm sorry, your question seems to be incomplete. Can you please provide more information or complete your question? Specifically, you mentioned "What volume of 0.05mol/dm3 NaCl," but you didn't mention what is the required quantity of NaCl you need to make this calculation. Any additional context you can provide will be helpful. Thank you.
How much hydroxide is required to make 25cm3of 0.1molar solution
To determine how much hydroxide is required to make a 25 cm3 solution of 0.1 M (mol/dm3) concentration, we first need to know the identity of the hydroxide.
Let's assume that the hydroxide is sodium hydroxide (NaOH), which is a common base.
From the formula of NaOH, we know that one mole of NaOH contains one mole of hydroxide ions (OH-).
To make 25 cm3 of a 0.1 M NaOH solution, we can use the following formula:
moles of NaOH = concentration (mol/dm3) x volume (dm3)
moles of NaOH = 0.1 mol/dm3 x 0.025 dm3
moles of NaOH = 0.0025 mol
Since one mole of NaOH contains one mole of hydroxide ions, the number of moles of hydroxide ions (OH-) required to make 25 cm3 of a 0.1 M solution is also 0.0025 mol.
Therefore, the amount of hydroxide required to make 25 cm3 of a 0.1 M solution depends on the identity of the hydroxide. For example, if we assume that the hydroxide is sodium hydroxide, then the amount of hydroxide required is 0.0025 mol or:
mass of NaOH = number of moles x molar mass
mass of NaOH = 0.0025 mol x 40.00 g/mol (molar mass of NaOH)
mass of NaOH = 0.1 g
So, 0.1 g of sodium hydroxide will be required to make a 25 cm3 solution of 0.1 M concentration.
A solution contains 0.1molar of the sodium per dm3 of solution .what volume of water in dm3 will be added to dilute it to 0.01g/dm3
To dilute a 0.1 M (mol/dm3) solution of sodium to 0.01 g/dm3, we need to calculate the new volume of the diluted solution after the addition of water. We can use the formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, we know that:
C1 = 0.1 M
C2 = 0.01 g/dm3
We can convert the final concentration in grams per dm3 to molarity using the molar mass of sodium (Na) which is 22.99 g/mol:
C2 = 0.01 g/dm3 ÷ 22.99 g/mol = 4.35 x 10^-4 M
We also know that V1 (the initial volume) is 1 dm3, since the solution contains 0.1 moles of sodium per dm3 of solution.
Now we can rearrange the formula to solve for the final volume (V2):
V2 = (C1 x V1) / C2
V2 = (0.1 M x 1 dm3) / (4.35 x 10^-4 M)
V2 = 229.9 dm3
Therefore, to dilute the 0.1 M solution of sodium to 0.01 g/dm3, we need to add approximately 229.9 dm3 of water.
If 5.00cmof 2.20mol/dm Na2CO3 of diluted water with distilled water to obtain 250 solution what is the concentrations of resulting solution
To calculate the concentration of the resulting solution, we can use the formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, we have:
C1 = 2.20 mol/dm3 (the concentration of the initial 5.00 cm3 of Na2CO3 solution)
V1 = 5.00 cm3 (the initial volume)
V2 = 250 cm3 (the final volume, after dilution)
We can rearrange the formula to solve for the final concentration (C2):
C2 = (C1 x V1) / V2
C2 = (2.20 mol/dm3 x 5.00 cm3) / 250 cm3
C2 = 0.044 mol/dm3
Therefore, the concentration of the resulting solution is 0.044 mol/dm3.