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ad-work [718]
3 years ago
9

if you are working with a radioactive source emitting particles of several kinds all positively charged not necessarily of the s

ame mass. how could you determine whether all particles have the same charge?
Chemistry
1 answer:
zlopas [31]3 years ago
3 0

Pass the charged particles at high velocity through an electric field.

The positively charged particles will deflect one way and the negatively charged particles will deflect the opposite way. The deflection will observe the Right-Hand Rule. The rate of detection will also depend on the size of the charged particle with heavier one being deflected more.


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4 b + 3 o2 → 2 b2o3 if 8 moles of b and 4 moles of o2 are allowed to react, how many moles of b2o3 can be formed?
raketka [301]

The number of moles of  b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

4 b + 3O₂ → 2b₂O₃

from the equation above;

4 b ------------> 2 b₂O₃

2b ------------> b₂O₃

2 : 1

3O₂  -------------> 2b₂O₃

3  :  2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of  b2o3

8 moles -------> ?

= (8 x 2)/4

= 4 moles

Thus, the number of moles of  b2o3 that will be formed is determined as 4 moles.

Learn more about limiting reactants here: brainly.com/question/14222359

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6 0
2 years ago
HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.
Ghella [55]

Answer:

\boxed{\text{66.95 g BaSO$_{4}$}}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:         261.34                         233.39

              Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g:         75.00

1. Moles of Ba(NO₃)₂

\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}

3. Mass of BaSO₄

\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}

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