Who first determined the value of the avogadros number

How many moles of oxygen are in 2.00 moles of NaCIO4?

Klio2033 [76]

Answer: 8 Moles of Oxygen

Explanation: The formula shows that there are (exactly) 4 oxygen atoms in each mole or formula unit. Therefore 4 * 2.00 = 8.

3 0

3 years ago

Calculate the specific heat in calories per °C of a 2.00 lb pure iron bar. Plz help me!!!

nekit [7.7K]

Answer:

402.7897

Explanation:

Pretty sure but not 100% about it.

grams in 2 pounds = 907.184 * specific heat of iron = .444

=402.789696

4 0

3 years ago

At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements

Gnom [1K]

Given that:

At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

Initial volume of Final volume of Mass of water Density (g/mL)

burette (mL) burette (mL) dispensed (g)

Sample 1 2.33 7.34 5.000 -----

Sample 2 7.34 12.37 5.025 -----

Sample 3 12.37 18.50 6.112 -----

Sample 4 18.50 24.57 6.064 -----

Sample 5 24.57 31.31 6.720 -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

$\mathbf{density = \dfrac{mass}{volume}}$

$\mathbf{density = \dfrac{5.01 g}{5.000 ml}}$

density = 0.998004 g/ml

Sample 2:

$\mathbf{density = \dfrac{5.025 g}{5.03ml}}$

density = 0.999006 g/ml

Sample 3:

$\mathbf{density = \dfrac{6.112 g}{6.13ml}}$

density = 0.997064 g/ml

Sample 4:

$\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}$

density = 0.999012 g/ml

Sample 5:

$\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}$

density = 0.997033 g/ml

Thus, the average density for all the samples is:

$\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}$

= 0.998024

∴

The percentage error for the two densities measurement is:

$=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}$

Given that the theoretical value = 0.997044 g/ml

Then;

$\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}$

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0

2 years ago

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

jeyben [28]

Answer: The amount of barium sulfate produced in the given reaction is 0.667 grams.

Explanation:

To calculate the number of moles from molarity, we use the equation:

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

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0

3 years ago

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago

Who first determined the value of the avogadros number​

Who first determined the value of the avogadros number