Answer:
because of the location of the 6 the answer is 6 cent
Answer:
31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .
Explanation:
Given data:
Number of moles of CaCl₂ = 2.6 mol
Number of Cl₂ ions = ?
Solution:
CaCl₂ → Ca²⁺ + 2Cl⁻
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
In one mole of CaCl₂ there are two moles of chloride ions present.
In 2.6 mol:
2.6×2 = 5.2 moles
1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions
5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol
31.31× 10²³ number of Cl⁻
Answer: A barrier should be created to overcome the atmosphere of the Venus, while launching spacecraft to Venus.
Explanation:
The atmosphere of Venus consists of 96.5% carbon dioxide, other composition includes nitrogen and other gases in trace amounts. The large amount of carbon dioxide in the atmosphere can extinguish the missile of the launcher of spacecraft thus it will become difficult in launch of spacecraft to the Venus.
Answer:
The volume of carbon dioxide gas generated 468 mL.
Explanation:
The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%
Mass of tablet = 3.45 g
Mass of bicarbonate =![3.45 g\times \frac{32.5}{100}=1.121 mol](https://tex.z-dn.net/?f=3.45%20g%5Ctimes%20%5Cfrac%7B32.5%7D%7B100%7D%3D1.121%20mol)
Moles of bicarbonate ion = ![\frac{1.121 g/mol}{61 g/mol}=0.01840 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1.121%20g%2Fmol%7D%7B61%20g%2Fmol%7D%3D0.01840%20mol)
![HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)](https://tex.z-dn.net/?f=HCO_3%5E%7B-%7D%28aq%29%2BHCl%28aq%29%5Crightarrow%20H_2O%28l%29%2BCO_2%28g%29%2BCl%5E-%28aq%29)
According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:
of carbon dioxide gas
Moles of carbon dioxide gas n = 0.01840 mol
Pressure of the carbon dioxide gas = P = 1.00 atm
Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K
Volume of the carbon dioxide gas = V
(ideal gas equation)
![V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D%3D%5Cfrac%7B0.01840%20mol%5Ctimes%200.0821%20atm%20L%2Fmol%20K%5Ctimes%20310%20K%7D%7B1.00%20atm%7D%3D0.468%20L)
1 L = 1000 mL
0.468 L =0.468 × 1000 mL = 468 mL
The volume of carbon dioxide gas generated 468 mL.
Answer: The formula mass (formula weight) of a molecule is the sum of the atomic weights of the atoms in its empirical formula. The molecular mass (molecular weight) of a molecule is its average mass as calculated by adding together the atomic weights of the atoms in the molecular formula.
Hope this helps.... Stay safe and have a great weekend!!!!! :D