1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.
Answer:
Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre.
Explanation:
Hydrocarbons are compounds formed by only hydrogen atoms and carbon.
Answer (2)
hope this helps!
Answer:
Methods of soil conservation are listed below.
Explanation:
The major sources of soil erosion include water,wind and tillage. In order to mitigate or prevent soil erosion, some of the following techniques can be implemented:
- <u>Contour Farming: </u>Planting in row patterns that run level around a hill — as opposed to the up and down the slope pattern.This reduces runoffs and consequently water erosion.
- <u>Crop Rotation:</u> This involves planting crops with high residue (e.g corn, small grains, e.t.c) in rotation,as the layer of residue would protect the topsoil.
- <u>Built in structural diversion</u> : Used often for gully control, to regulate flow of water away from the field and through designated desired paths.
- <u>Conservation Tillage</u>: This involves methods such as no-till planting, strip rotary tillage, etc, which do not allow the soil surface to be smooth and bare, but instead covered with crop residue that protects the soil from eroding forces.
Answer:
The new volume of the balloon will be 6046.28 L
Explanation:
Initial pressure (P1) = 99 kpa
initial volume (V1) = 3000 L
Initial temperature = 39 C = 39 + 273 = 312 K
Final pressure (P2) = 45.5 kpa
Final temperature = 16 C = 16 +273 = 289K
Final volume = ????
To calculate the final volume using the general gas equation
P1 V1 / T1 = P2 V2 / T2
make V2 the subject of the formular
V2 = 99000 ×3000× 289 / 45500×312
V2 = 85833000 /14196
V2 = 6046.28 litres
