Answer:
Density of iron shot is 7.92 g/mL.
Explanation:
Given data:
Mass of water + iron + graduated cylinder = 94.74 g
Mass of water + graduated cylinder = 66.24 g
Volume of water = 45.50 mL
Volume of water + iron shot = 49.10 mL
Density of iron = ?
Solution:
First of all we will calculate the volume of iron shot.
Volume of iron shot= (Volume of water + iron shot) - Volume of water
Volume of iron shot= 49.10 mL - 45.50 mL
Volume of iron shot= 3.6 mL
Now we will calculate the mass of iron shot.
Mass of iron shot = (Mass of water + iron + graduated cylinder ) - Mass of water + graduated cylinder
Mass of iron shot = 94.74 g - 66.24 g
Mass of iron shot = 28.5 g
Now we will calculate the density of iron shot.
d = m/v
d = 28.5 g / 3.6 mL
d = 7.92 g/mL
Density of iron shot is 7.92 g/mL.
Fe + 3NaBr → FeBr3 + 3Na
The Na is replaced by the Fe atom.
Answer:
15.0 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In this equation,
-----> P = pressure (mmHg)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)
-----> T = temperature (K)
To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams
P = 728 mmHg R = 62.36 L*mmHg/mol*K
V = ? L T = 45.0 °C + 273.15 = 318.15 K
n = 0.551 moles
PV = nRT
(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)
(728 mmHg)V = 10922.7632
V = 15.0 L
Answer:
Different forces between atoms and difference in structure.
7.66g of ethyl butyrate is produced.
The reaction is
CH₃CH₂CH₂COOH + CH₃CH₂OH ----> CH₃CH₂CH₂COOCH₂CH₃
- The molar mass of butanoic acid is 88.11g/mol
- We have 7.45g of butanoic acid
- The moles of butanoic acid we have is 7.45/88.11 = 0.0845 mol
- If the yield is 100%, 1 mole of butanoic acid gives 1 mole of ethyl butyrate
- But the reaction yield is 78%
- 1 mole of butanoic acid gives 0.78 mole of ethyl butyrate
From 0.0845 mol of butanoic acid we get 0.78 x 0.0845 = 0.66 mol of ethyl butyrate.
The molar mass of ethyl butyrate is 116.16g/mol
So 0.66 x 116.16 = 7.66g
7.66g of ethyl butyrate is produced.
Learn more about esterification here:
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