All categories

3 years ago

When was the first published textbook containing a table identifying the first 33 elements

1 answer:

5 0

1789 -- raité élémentaire de chimie is a textbook written by Antoine Lavoisier published in 1789 and translated into English by Robert Kerr in 1790 under the title Elements of Chemistry

You might be interested in

Calculate the concentration imports per million ppm of DDT if a sample size of 5000 g contained 0.10 g DDT

nevsk [136]

Answer:

20ppm

Explanation:

parts per million are defined as the mass of solute in mg (In this case, mass of DDT) per kg of sample.

To solve this question we must find the mass of DDT in mg and the mass of sample in kg:

0.10g * (1000mg / 1g) = 100mg

<em>Mass sample:</em>

5000g * (1kg / 1000g) = 5kg

Parts per Million:

100mg / 5kg =

3 0

3 years ago

Which of the five most abundant elements are metals?

Gala2k [10]

Answer:

Oxygen

Silicon

Aluminium

Iron

Calcium

Explanation:

Oxygen=47 percent

Silicon=28 percent

Aluminium=8 percent

Iron=5 percent

Calcium=3.5 percent

Hope it helps :)

Mark me as brainliest

6 0

3 years ago

202H8(9) + 7 O2(g) > 4CO2(g) + 6 H2O(9).

lilavasa [31]

since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.

4 0

3 years ago

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of $AgNO_{3}$ in 50.0 mL of 0.100 M of $AgNO_{3}$

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= $\frac{0.100}{1000}\times 50.0$ moles

= 0.00500 moles

According to balanced equation, 1 mol of $AgNO_{3}$ reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of $AgNO_{3}$ react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = ( $100.0\times 1.00$ ) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= $\frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}$

= $\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}$

= -67716 J/mol

[m = mass, c = specific heat capacity, $\Delta T$ = change in temperature and negative sign is included as it is an exothermic reaction]

4 0

3 years ago

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.

7 0

3 years ago