The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer:
The symbol of the ion is A^-
Explanation:
Let A be the symbol of the element.
Proton = 17
Electron = 18
Neutron = 20
Since the element has more electrons than protons, it means it has gain electrons.
From the question given,
The difference between the electron and proton = 18 — 17 = 1
So, the element has gain 1 electron.
Therefore the symbol of the ion is A^-
Answer:
1.1 × 10²⁴ atoms Mg
General Formulas and Concepts:
<u>Atomic Structure</u>
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] 1.8 mol Mg
[Solve] atoms Mg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
Velocity = 15 m/s
90/2 = 45 m
45/3 (seconds to reach) = 15 m