Answer:
14.93 g
Explanation:
First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:
- 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As
Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:
- 0.199 mol * 74.92 g/mol = 14.93 g
Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.
Reactant C is the limiting reactant in this scenario.
Explanation:
The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.
Balanced chemical reaction is:
A + 2B + 3C → 2D + E
number of moles
A = 0.50 mole
B = 0.60 moles
C = 0.90 moles
Taking A as the reactant
1 mole of A reacted to form 2 moles of D
0.50 moles of A will produce
= 
thus 0.50 moles of A will produce 1 mole of D
Taking B as the reactant
2 moles of B reacted to form 2 moles of D
0.60 moles of B reacted to form x moles of D
= 
x = 2 moles of D is produced.
Taking C as the reactant:
3 moles of C reacted to form 2 moles of D
O.9 moles of C reacted to form x moles of D
= 
= 0.60 moles of D is formed.
Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.
a. the ratio of mass to charge of an electron
Explanation:
The experiment permitted the direct measurement of the ratio of mass to charge of an electron.
- The charge to mass ratio of an electron was determine by accelerating a beam of cathode rays in magnetic and electric fields.
- No matter the gas used in the tube or the nature of the material of the electrodes, the rays were found to have constant charge to mass ratio of 1.76 x 10¹¹coulombkg⁻¹.
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Answer: none
Explanation: there is no picture attached
Answer : The correct option is, 
Explanation :
To calculate the pressure of gas we are using ideal gas equation as:

where,
P = pressure of gas = ?
V = volume of gas = 0.046 L
n = number of moles of gas = 3.4
R = gas constant = 8.314 L.kPa/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:


Therefore, the pressure of gas is, 