All categories

3 years ago

I need to know which piece of research in the square was done by John Dalton.

Chemistry

1 answer:

sashaice [31]3 years ago

8 0

Answer:

C. atomic mass

Explanation:

John Dalton was a very versatile man. His work is important in physics, meteorology, medicine (description of color blindness) and chemistry. He was included in researching of partial pressures and atomic theory. He proposed one of the first models of atom in the beginning of 19th century.

He published a table where he proposed relative atomic weight of six atoms. In his honor, standardized unit for atomic weight is called dalton (Da) and weighs 1/12 of carbon atom weight.

You might be interested in

Classifying Animals

MrMuchimi

Answer:

I believe:

vertebrate: hummingbird, tiger, lobster

invertabrate: lizard, octopus, mosquito

4 0

3 years ago

Read 2 more answers

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

3 0

3 years ago

Rare earth elements plz

Slav-nsk [51]

Rare earth metals are a group of 17 elements - lanthanum, cerium, praseodymium, neodymium, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, lutetium, scandium, yttrium - that appear in low concentrations in the ground

3 0

3 years ago

Read 2 more answers

help asap show work | a gas with a volume of 13L at a pressure of 197 kPA is allowed to expand to a volume of 23L. what is the p

Lera25 [3.4K]

Answer:

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$