Na - 1s^2, 2s^2, 2p^6, 3s^1
Ne - 1s^2, 2s^2, 2p^6
O - 1s^2, 2s^2, 2p^4
When y equals 5, x is 104.3
When y equals 3 then x is 108.3
<em><u>Solution:</u></em>
<em><u>Given expression is:</u></em>
<h3><u>If y equals 5 what is x ?</u></h3>
Substitute y = 5 in given expression
5 = 57.15 - 0.5(x)
5 = 57.15 - 0.5x
0.5x = 57.15 - 5
0.5x = 52.15
Divide both sides by 0.5
x = 104.3
Thus when y equals 5, x is 104.3
<h3><u>If y = 3 what is x ?</u></h3>
Substitute y = 3 in given expression
3 = 57.15 - 0.5(x)
3 = 57.15 - 0.5x
0.5x = 57.15 - 3
0.5x = 54.15
Divide both sides by 0.5
x = 108.3
Thus when y equals 3 then x is 108.3
Since there is no phase change, we can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J kg⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 2000 J
m = 100 g = 0.1 kg
c = ?
ΔT = (70 °C - 50 °C) = 20 °C
By applying the formula,
2000 J = 0.1 kg x c x 20 °C
c = 2000 J / (0.1 kg x 20 °C)
c = 1000 J kg⁻¹ °C⁻¹
Hence, the specific heat capacity of the liquid is 1000 J kg⁻¹ °C⁻¹.
Answer:
The mass of copper sulfate in 25 cm³ of the copper sulfate solution is 0.45 g
Explanation:
The given parameters are;
The volume of the copper sulfate solution = 100 cm³
The mass of the copper sulfate in the solution = 1.8 g
Therefore, the mass of copper sulfate in 25 cm³ of the solution is given as follows;
The mass of copper sulfate in 100 cm³ of the solution = 1.8 g
The mass of copper sulfate in 1 cm³ of the solution = 1.8 g/100 = 0.018 g
Therefore;
The mass of copper sulfate in 25 × 1 = 25 cm³ of the solution, m = 25×0.018 g = 0.45 g
∴ The mass of copper sulfate in 25 cm³ of the solution, m = 0.45 g
Answer:
Explanation:
<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
- Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C
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<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
- proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H
<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>
- Mass of O = mass of pure sample - mass of C - mass of H
- Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O
Round to four decimals: Mass of O = 0.3840 g
<u>4) Mole calculations</u>
Divide the mass in grams of each element by its atomic mass:
- C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
- H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
- O: 0.3840 g / 15.999 g/mol = 0.02400 mol
<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>
- C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
- H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
- O: 0.02400 mol / 0.02400 mol = 1
Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:
