<span> are composed of the fragments, or CLASTS. If PRE-existing </span>minerals<span> and rock. A </span>clast<span> is a fragment of </span>geological detritus,<span>chunks and smaller grains of rock broken off other rocks by </span>physical weathering.[2]<span> Geologists use the term CLASTIC </span><span>with reference to </span>sedimentary rocks<span> as well as to particles in </span>sediment transport<span> whether in </span>suspension<span> or as </span>bed load<span>, and in </span>sediment<span> deposits.</span>
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
Answer:
CH3COONa : CH3COOH = 1.74 : 1 (required to prepare a buffer at PH = 5)
Explanation:
Given that :
PH = 5
Calculate proportions of ethanoic acid and Sodium ethanoate to prepare Buffer solution
solution :
PH = Pka + Log ( CH3COONa / CH3COOH )
5.0 = 4.76 + Log ( CH3COONa / CH3COOH )
∴ Log ( CH3COONa / CH3COOH ) = 0.24
( CH3COONa / CH3COOH ) = 1.74
Therefore : CH3COONa : CH3COOH
1.74 : 1
Answer:
445 mL NH₄Br
Explanation:
(Step 1)
Convert grams ammonium bromide (NH₄Br) to moles using its molar mass.
Molar Mass (NH₄Br): 14.007 g/mol + 4(1.008 g/mol) + 79.904 g/mol
Molar Mass (NH₄Br): 97.943 g/mol
8.42 grams NH₄Br 1 mole
----------------------------- x ------------------------ = 0.0860 moles NH₄Br
97.943 grams
(Step 2)
Calculate the volume of NH₄Br using the molarity equation. Then, convert liters (L) to milliliters (mL).
Molarity (M) = moles / volume (L)
0.193 M = 0.0860 moles / volume
(0.193 M) x volume = 0.0860 moles
volume = 0.445 L
1,000 mL = 1 L
0.445 L NH₄Br 1,000 mL
------------------------ x ------------------ = 445 mL NH₄Br
1 L
