Which acid is the acid usually found in your stomach?

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

3 years ago

In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

Answer: The moles of hydroxide ions present in the sample is 0.0008 moles

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

2 years ago

Limitation for osmosis experiment

taurus [48]

What was the question here?

5 0

1 year ago

Some nonliving things that is found in earths environment use an example of how you use each thing

Gnoma [55]

Rocks, rocks can be used for many things depending on which rock you are talking about some rocks are used for construction of cement mixtures. Granite and marble are used for counter tops and desks depending on what district you go to school in and how funded the schools are there. pumice is a rock found in waters around volcanoes and is used to smooth organic surfaces such as feet. miscellaneous rocks are also used to prevent erosion. non living things can also include carbon which is the bases of life itself. Carbon is an organic element which is used in the synthesis medicines such as aspirin and poisons such as dichloromethane which is not used to poison others but is toxic to humans, it is mainly used for different isolation of unknown substances. Depending on what R grouping is attached to the carbon can be the deciding factor of how it can be used and how it can react with the environment. Elements like helium can be used in gas chromatograms and put into balloons

6 0

3 years ago

How many moles of gas occupy 56.3 L at 0.899 atm and 20.0°C?

zzz [600]

Answer:

n = 2.1 mol

Explanation:

Given data:

Number of moles of gas = ?

Volume of gas = 56.3 L

Pressure of gas = 0.899 atm

Temperature of gas = 20°C (20+273 = 293 k)

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

0.899 atm × 56.3 L = n × 0.0821 atm.L/ mol.K × 293 k

50.614 atm.L = n × 24.055 atm.L/ mol

n = 50.614 atm.L / 24.055 atm.L/ mol

n = 2.1 mol

3 0

3 years ago