Answer:
answers are in attachment
Explanation:
1. Synthesis Reaction: In this type of reaction multiple reactant combine to form a single product.
2. Decomposition Reaction: In this type of reaction single compound or reactant break down into new elements or compounds.
3. Replacement Reaction: In this type of reaction an element replace an element in a compound.
4. Even hydrogen is not a metal but it can act as metal in single replacement reaction.
5. In combustion reaction fuel react with oxygen and give heat and light that increase temperature of surrounding.
6. In in-complete combustion reaction fuel react with in-sufficient oxygen and give carbon monoxide and carbon in form of soot.
7. Base is a compound that liberate OH⁻ ion in water.
8. indicators have different color in acid and base depends on pH of the solution
9. Salt and water are the product of neutralization reaction
NaOH + HCl = NaCl + H₂O
10. The pH of neutral solution is 7.
Answer:
The answer to your question is V = 0.32 L
Explanation:
Data
Volume of NH₃ = ?
P = 3.2 atm
T = 23°C
mass of CaH₂ = 2.65 g
Balanced chemical reaction
6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂
Process
1.- Convert the mass of CaH₂ to moles
-Calculate the molar mass of CaH₂
CaH₂ = 40 + 2 = 42 g
42 g ------------------ 1 mol
2.65 g -------------- x
x = (2.65 x 1)/42
x = 0.063 moles
2.- Calculate the moles of NH₃
2 moles of NH₃ --------------- 3 moles of CaH₂
x --------------- 0.063 moles
x = (0.063 x 2) / 3
x = 0.042 moles of NH₃
3.- Convert the °C to °K
Temperature = 23°C + 273
= 296°K
4.- Calculate the volume of NH₃
-Use the ideal gas law
PV = nRT
-Solve for V
V = nRT / P
-Substitution
V = (0.042)(0.082)(296) / 3.2
-Simplification
V = 1.019 / 3.2
-Result
V = 0.32 L
Answer:
FALSE
Explanation:
false. kinematics equations are used for many purposes.it is used to drive equations and to find the motion of an object. but it is used more in physics rather than maths.
The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.
Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be 
g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.
