All categories

2 years ago

The weights of all elements are always compared to oxygen.

Chemistry

1 answer:

marishachu [46]2 years ago

7 0

Answer:

The weights of all elements are always compared to the Carbon-12.

Explanation:

The weights of all elements are always compared to the Carbon-12 because the mass of carbon is 12 which is the exactly the sum of protons and neutrons.

Oxygen was also considered the standard for some time but later this stander was rejected because in natural O¹⁷ and O¹⁸ were also present and this create the two different atomic mass tables.

AMU:

Atomic mass unit is define as the 1/12 the mass of an atom of carbon-12.

C12 has six neutron and six protons in the nucleus.

This unit is used to express the masses of atoms. We know that masses of atoms are very small and we do not have any such type of balance that can measure very small quantity. So that is way we use this scale to measure small quantity. For example, according to this scale

relative atomic mass of hydrogen is 1.008 amu

relative atomic mass of oxygen is 15.999 amu

relative atomic mass of uranium is 238.0289 amu

relative atomic mass of chlorine is 35.453 amu

You might be interested in

- Nitinol is an alloy of *

Aleks04 [339]

Answer:

Nickel and Titanium

Explanation:

Nitinol is an alloy of Nickel and Titanium. It posesses two properties such that,

The shape memory effect
Super elasticity

Shape memory is the ability of nitinol to undergo deformation at one temperature, stay in its deformed shape when the external force is removed.

Superelasticity is the ability for the metal to undergo large deformations and immediately return to its undeformed shape upon removal of the external load.

Hence, the correct option is (b) "Nickel and Titanium".

7 0

3 years ago

When a compound such as X2Z4 is created, what would be the<br> correct simplified chemical formula?

Shtirlitz [24]

Answer:

XZ2

Explanation:

There are different ways in which compounds can be represented. Broadly, we have three different types of formula;

- Structural formular: This shows how th atoms in te compound are connected to each other.

- Molecular formular: This shows the actual number of atoms of element present in the compound

- Empirical Formular: This is the simplest formular of a compound. It basically shows the number of atoms in simple ration to each other.

This question requires us to input the empirical formular;

X2Z4

The ratio of the elements is; 2 : 4 which can be simplified into 1 : 2

This means the empirical formular is XZ2

6 0

2 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

2 years ago

HC]A laboratory experiment requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M

andreyandreev [35.5K]

The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.

3 0

2 years ago

If I have 2 liters of gas held at a pressure of 78 atm and a temperature of 900 k, what will be the volume of the gas if I decre

dangina [55]

Answer:

V2 = 2.88L

Explanation:

P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K

Applying the general gas equation

P1V1/T1 = P2V2/T2

Substitute the above

(78*2)/900= (45*V2)/750

V2= (78*2×750)/(900*45)

V2= 2.88L

7 0

3 years ago