Answer:
Substance is Aluminium
Explanation:
We are given;
Mass; m = 13.5 grams
Volume; V = 5 cm³
Formula for density is;
density = m/V
Density = 13.5/5
Density = 2.7 g/cm³
From the table attached, we can see that the element with a corresponding density of 2.7 is Aluminium
Answer:
B. 1.65 L
Explanation:
Step 1: Write the balanced equation
2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)
Step 2: Calculate the moles of SO₂
The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol
Step 3: Calculate the moles of SO₃ produced
0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃
Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
0.0736 mol × 22.4 L/1 mol = 1.65 L
Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
Taking the average of more measurements decreases random error of measurement
Taking the average of many measurements is the most effective way to reduce random errors in a measurement. Because the certainty of the results grows as the number of data does, Less risk of random errors means that the value is more certain. Fewer measurements lead to less reliable data collection, which raises the likelihood of random errors.
The complete question is
Which procedure(s) decrease(s) the random error of a measurement: (1) taking the average of more measurements: (2) calibrating the instrument; (3) taking fewer measurements? Explain
To learn more about random errors:
brainly.com/question/14149934
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