Which of the following bases can remove a proton from formic acid in a reaction that favors products?

It take 38.70cm³ of 1.90m NaoH to neutralize 10.30cm³ of H2so4 in a battery, calculate the molar concentration of H2so4

zlopas [31]

Answer:

$M_{acid}=3.57M$

Explanation:

Hello there!

In this case, since this acid-base neutralization is performed in a 1:2 mole ratio of acid to base as the former is a diprotic acid (two hydrogen ions in the molecule), we can write the following equation:

$2M_{acid}V_{acid}=M_{base}V_{base}$

In such a way, we can solve for the molarity of the acid, given the molarity and concentration of the NaOH base and the volume of the acid:

$M_{acid}=\frac{M_{base}V_{base}}{2V_{acid}}$

Thus, we plug in the given data to obtain:

$M_{acid}=\frac{38.70cm^3*1.90M}{2(10.30cm^3)} \\\\M_{acid}=3.57M$

Best regards!

8 0

2 years ago

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a

geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = 0.182 grams

2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0

2 years ago

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon

Ulleksa [173]

Answer:

Theoretical yield = 31.8 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $Al_2O_3$

Mass of $Al_2O_3$ = 60.0 g

Molar mass of $Al_2O_3$ = 101.96128 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{60.0\ g}{101.96128\ g/mol}$

$Moles_{Al_2O_3}= 0.5885\ mol$

Given: For $C$

Given mass = 30.0 g

Molar mass of $C$ = 12.0107 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{30.0\ g}{12.0107\ g/mol}$

$Moles_{C}= 2.4978\ mol$

According to the given reaction:

$Al_2O_3+3C\rightarrow 2Al+3CO$

1 mole of aluminium oxide react with 3 moles of carbon

0.5885 mole of aluminium oxide react with $3\times 0.5885$ moles of carbon

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

Limiting reagent is the one which is present in small amount. Thus, aluminium oxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of aluminium oxide on reaction forms 2 moles of aluminium.

0.5885 mole of aluminium oxide on reaction forms $2\times 0.5885$ moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

Theoretical yield = 31.8 g

3 0

2 years ago

A solution is prepared by dissolving 91.7 g fructose in 545 g of water. Determine the mole fraction of fructose if the final vol

oee [108]

Answer:

Mole fraction = 0,0166

Explanation:

Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.

The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:

91,7g fructose × (1mol / 180,16g) = 0,509 moles of fructose

545g water × (1mol / 18,02g) = 30,24 moles of water

Thus, mole fraction of fructose is:

$\frac{0,509 moles}{0,509mol + 30,24mol} = 0,0166$

Mole fraction = 0,0166

I hope it helps!

5 0

2 years ago

What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the

bulgar [2K]

0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
So moles of protons = 0.01 x 2 = 0.02 moles of H+ 
For neutralization: moles H+ = moles OH- 
Therefore moles of NaOH = 0.02 
conc = moles / volume 
Conc NaOH = 0.02 / 0.025L = 0.8M

4 0

2 years ago

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