Answer:
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Answer:
A. 0.65g/mL, float
Explanation:
Mass: 325g
Volume: 500mL
D = m/v
D = 325/500
D = 0.65
Therefore, the block has a density of 0.65g/mL and it will float since the density of water is greater than the density of the block.
Answer:
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With determination and hard work, people can overcome difficult challenges. How do the people of Greensburg, Kansas, and Chicago, Illinois, illustrate this idea? Use details from both texts to support your response. i cant do it
Explanation:
Answer:
- g) 0.20 mol
- h) 2.0 mol/dm³
- i) 2.8 g
Explanation:
<h2>a) </h2>
1. Data:
2. Solution:
i) Formulae:
- number of moles = mass in grams/Ar
- number of atoms = number of moles × Avogadro constant
- Avogadro constant = 6.022 × 10²³ atoms/mol
ii) Calculations:
- number of moles = 71g / 35.5(g/mol) = 2 mol
- number of atoms = 2mol × 6.022 × 10²³ atoms/mol = 1.20 × 10²⁴ atoms
<h2>b) </h2>
1. Data:
- 0.2 mol KOH
- Ar: H = 1, O = 16, k = 39
2. Solution
i)Formula:
- mass = number of moles × molar mass
ii) Molar mass:
- 1×1g/mol + 1×39g/mol + 1×16g/mol = 56g/mol
iii) Calculations:
- mass = 0.2 mol × 56g/mol = 11.2 g/mol ≈ 10g/mol (rounded to 1 significant figure)
<h2>c) </h2>
1. Data:
2. Solution
i) Formula:
- number of moles = mass in grams / Ar
ii) Calculations:
- number of moles = 1.4g / 7 gmol = 0.7 mol
<h2>d) </h2>
1. Data:
2. Solution
i) Formula:
- mass = number of moles × molar mass
ii) Calculations:
- molar mass = 8 × 32g/mol = 256 g/mol
- mass = 4 mol × 256 g/mol = 1,024 g ≈ 1,000 g (rounded to 1 significant figure)
<h2>e)</h2>
1. Data:
- Mr: Ca(NO₃)₂ . 2H₂O = 200
2. Solution
i) Formulae:
- number of moles = mass in grams / molar mass
ii) Calculations:
- number of moles = 50 g / 200 g/mol = 0.25 mol of Ca(NO₃)₂ . 2H₂O
The number of moles of water molecules is 2 times the number of moles of Ca(NO₃)₂ . 2H₂O.
- number of moles of water molecules = 2 × 0.25 mol = 0.5 mol
<h2>f) </h2>
1. Data:
2. Solution:
i) Formulae:
- number of atoms = number of moles × Avogadro constant
- number of moles = mass in grams / molar mass
ii) Calculations
- number of moles = 4.9g / 98g/mol = 0.050 mol
- number of atoms = 0.050 mol × 6.022 × 10²³ atoms/mol = 3.0×10²² atoms
<h2>g) </h2>
1. Data:
- V = 24 dm³ (air)
- 20% oxygen
- r.t.p ⇒ T = 298K, p = 1 atm
2. Solution
i) Formula:
ii) Calculation:
- R = 0.08206 (atm.dm³/K.mol)
- n = [ 1 atm × 24 dm³] / (0.08206atm.dm³/K.mol × 298K) = 0.98mol of air
- moles of oxygen = 20% × 0.98 mol ≈ 0.20 mol
<h2>h) </h2>
1. Data:
2. Solution
i) Formula:
- Molarity = moles of solute / volume of solution in dm³
ii) Calculations:
25cm³ × (0.1cm/dm)³ = 0.125 dm³
- Molarity = 0.25 mol / 0.125 dm³ = 2.0 mol/dm³
<h2>i) </h2>
1. Data:
- V = 35cm³
- M = 2 mol/dm³
- Mr NaOH = 40
2. Solution
i) Formulae:
- number of moles = M × V (in dm³)
- mass = number of moles × mola rmass
ii) Calculations:
- V = 35cm³ × (0.1dm/cm)³ = 0.035dm³
- number of moles = 2 mol/dm³ × 0.035 dm³ = 0.070 mol
- mass = 0.070 mol × 40 g/mol = 2.8 g
Answer:
the energy required is Q = 4.9 kJ
Explanation:
Since the energy required is
Q = L * n
where L = molar enthalpy of fusion of solid sodium = 2.60 kJ/mol
and n = number of moles involved
also we know that
n= m/ M , where m= mass involved , M = molecular weight of sodium
the molecular weight of sodium is M= 23 g / mol
therefore
Q = L * n = L* m / M
replacing values
Q = L* m / M = 2.60 kJ/mol * 43.6 g/ (23 g / mol) = 4.9 kJ
Q = 4.9 kJ
thus the energy required is Q = 4.9 kJ