Answer:
It is
D- <u><em>DEFINE</em></u> the problem
R-<em><u>RESEARCH</u></em> on the problem
H- Carry out a <u><em>HYPOTHESIS</em></u>
E- carry out an <em><u>EXPERIMENT</u></em>.
R-Analyse the <u><em>RESULT</em></u>
C-summarise the <u><em>CONCLUSION</em></u>.
Explanation:
Hope it helps.
A chemical reaction must occur for a compound to have different properties.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
10.945 x 10^-4
Explanation:
Balanced equation:
Mn(OH)2 + 2 HCl --> MnCl2 + H2O
it takes 2 moles HCL for each mole Mn(OH)2
Next find the molarity of the Mn(OH)2 solution
= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)
= 4.86 x 10^-3 mole
this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L
thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M
Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
