<h3>
Answer:</h3>
3.01 × 10²⁵ molecules H₂O
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
50.0 mol H₂O
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 3.011 × 10²⁵ molecules H₂O
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O
Osmosis is the diffusion of water from a high concentration to a lower concentration
Answer : The mass of 
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of 
.
Now we have to calculate the moles of .
The balanced chemical reaction will be,
![PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4](https://tex.z-dn.net/?f=PbSO_4%2BNa_2CrO_4%5Crightarrow%20PbCrO_4%2BNa_2SO_4%5Btex%5D%3C%2Fp%3E%3Cp%3EFrom%20the%20balanced%20chemical%20reaction%2C%20we%20conclude%20that%3C%2Fp%3E%3Cp%3EAs%2C%201%20mole%20of%20%5Btex%5DPbCrO_4)
produced from 1 mole of
So, 0.005 mole of produced from 0.005 mole of
Now we have to calculate the mass of
Therefore, the mass of needed are, 1.515 grams.
Answer:
Their positive charge is located in the small nucleus
Explanation:
Ernest Rutherford performed the gold foil experiment in 1911 where he used alpha particles generated from a radioactive source to bombard a thin gold foil.
In his experiment, he observed that the bulk of the alpha particles passed through the gold foil, just a tiny fraction was deflected back. To explain his findings, Rutherford proposed that an atom is made of positively charged centre where nearly all the mass is concentrated called nucleus. Surrounding the nucleus is a large space containing electrons.
Answer:
It kinda helps but not really
Thanks for trying anyway doe!
Explanation:
