Answer:
Mostly large grains, with a sticky texture, 55% sand, 40% clay, and 5% silt
Explanation: I took the test, hope it helps.
Answer: 4.15234 m
512 g H2O * 
= 0.512 kg H2O
Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol
H = 1.008 g/mol
N = 14.007 g/mol
O3 = 3*15.999
134 g HNO₃ * 
= 2.126 mol
m = 
= 4.15234 m
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
(1) Ploar covalent. is the answer.
