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Vsevolod [243]
3 years ago
11

Why is Avogadro’s number referred to as a mole? If you could change this terminology, what would you change it to and why?

Chemistry
1 answer:
stepan [7]3 years ago
7 0

The number of particles (molecules, atoms, compounds, etc.) per mole of a substances is known as Avagadro number. It is equal to 6.022×10^23 mol-1 and is expressed as NA.


Number of moles is the amount of a substance that contains as many particles as there are atoms in 12 grams of pure carbon-12. So, 1 mol contains 6.022×10^23 elementary entities of the substance. Since 6.022 x 10^23 is the Avagadro number, one mole is equal to Avagadro number.


One mole of a substance is the ratio of mass of the substance by the molecular mass of the substance. Thus the mass of one mole of a substance is equal to the substance's molecular weight. Thus one mole of a substance is the atomic mass unit of a substance and since one mole is equivalent to the Avagadro number,we can conclude that one Avagadro number is one atomic mass unit of the substance.

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What mass of magnesium would combine with exactly 16.0 grams of oxygen
TEA [102]
1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO

solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16

24.6 g is the answer.

if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.
3 0
3 years ago
A saturated solution of baso4 has a concentration of 0.5mol/l. a 55ml sample is taken by you. what is the mass of baso4 in the s
SIZIF [17.4K]

Answer:

6.4 g BaSO₄

Explanation:

You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.

Molarity (mol/L) = moles / volume (L)

(Step 1)

55 mL / 1,000 = 0.055 L

Molarity = moles / volume                             <----- Molarity ratio

0.5 (mol/L) = moles / 0.055 L                        <----- Insert values

0.0275 = moles                                             <----- Multiply both sides by 0.055

(Step 2)

Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (BaSO₄): 233.387 g/mol

0.0275 moles BaSO₄          233.387 g
---------------------------------  x  -------------------  =  6.4 g BaSO₄
                                                1 mole

6 0
2 years ago
An element in group 7 on the periodic table most likely has which physical property?
larisa86 [58]

Answer:

They are most likely solid

Explanation:

solid is a physical property

3 0
3 years ago
Is KNO a strong acid?
stira [4]

Answer:

First of all, it's KNO₃ not KNO.

Second, KNO₃ is neither an acid nor it is a base, infact, it is a salt and therefore it's neutral.

hope that helps...

3 0
2 years ago
A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it
Effectus [21]

Answer:

The pressure in the new flask would be 128\; \rm mmHg if the \rm HCl here acts like an ideal gas.  

Explanation:

Assume that the \rm HCl sample here acts like an ideal gas. By Boyle's Law, the pressure P of the gas should be inversely proportional to its volume V.

For example, let the initial volume and pressure of the sample be V_1 and P_1. The new volume V_2 and pressure P_2 of this sample shall satisfy the equation: P_1 \cdot V_1 =P_2 \cdot V_2.

In this question,

  • The initial volume of the gas is V_1= 256\; \rm mL.
  • The initial pressure of the gas is P_1 = 67.5\; \rm mmHg.
  • The new volume of the gas is V_2 = 135\; \rm mL.

The goal is to find the new pressure of this gas, P_2.

Assume that this sample is indeed an ideal gas. Then the equation P_1 \cdot V_1 =P_2 \cdot V_2 should still hold. Rearrange the equation to separate the unknown, P_2. Note: make sure that the units for V_1 and V_2 are the same before evaluating. That way, the unit of

\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}.

3 0
3 years ago
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