Answer:
A. 2 sig figs
Explanation:
In the number 0.0034, the zeros are called leading zeros. The only function of leading zeros is to fix the decimal point. Leading zeros are not counted as significant figures.
Explanation:
First, we need to calculate the number of moles of sodium carbonate we have in a 25 g sample. To calculate this, we will
find the molar mass of sodium carbonate (Na2CO3):
⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen
⇒ 2 × 23 + 12 + 3 × 16
⇒ 46 + 12 + 48
⇒ 106g/mol
Thus, the molar mass of Na2CO3 is 106g/mol.
Therefore, number of moles = 25 ÷ 106
=> 0.2358 mol
Now, we know that every mole of Na2CO3 have 0.2358 moles of Na+ ions. Hence, total moles of Na2CO3 is 0.4716 moles
Number of ions present = 6.022 × 1023 × 0.4716 mol = 2.84 × 1023ions
C. Decreasing the temperature
D. Raising the pressure
<h3>Further explanation</h3>
Given
Reaction
2SO₂+O₂⇔2SO₃+energy
Required
Changes to the formation of products
Solution
The formation of SO₃ is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature decreases, the equilibrium will shift towards the exothermic reaction, so the reaction shifts to the right towards SO₃( products-favored)
And increasing the pressure, then the reaction shifts to the right SO₃( products-favored)⇒the number of coefficients is greater
Answer:
1.20 V
Explanation:
The standard cell potential is calculated from the expression
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction
The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.
Thus for our question we will have
oxidation:
Pb(s) → Pb2+(aq) + 2 e- ε⁰ oxidation = - ε⁰ reduction
= - ( - 0.13 V ) = + 0.13 V
reduction
Br2(l) + 2 e- → 2 Br-(aq) ε⁰ reduction = +1.07 V
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction = + 0.13 V + 1.07 V = 1.20 V
Surface area<span> of a solid object is a measure of the total </span>area<span> that is occupied by the surface of object.
If two objects has same mass, then smaller size particles will have greater surface area.
Thus, in order to increase the surface area of ice, one can crush it into small particles. Due to crushing, particles will break into smaller size and eventually this will result in larger surface area. </span>
