We know that ;
Force = mass × acceleration
=> Force = 3×3 = 9N
_________
hope it helps!
Answer:
455.4 g
Explanation:
Data given:
no. of moles of (NH₄)₂SO₄= 3.45 mol
mass of (NH₄)₂SO₄ = ?
Solution
Formula will be used
no.of moles = mass in grams / molar mass
Rearrange the above equation for mass
mass in grams = no. of moles x molar mass . . . . . . . . (1)
molar mass of (NH₄)₂SO₄
molar mass of (NH₄)₂SO₄ = 2(14 + 4(1)) + 32 + 4(16)
molar mass of (NH₄)₂SO₄ = 2 (14 +4) + 32 + 64
molar mass of (NH₄)₂SO₄ = 2 (18) + 32 + 64
molar mass of (NH₄)₂SO₄ = 36 + 32 + 64 = 132 g/mol
Put values in equation 1
mass in grams = 3.45 mole x 132 g/mol
mass in grams = 455.4 g
So,
mass of (NH₄)₂SO₄ = 455.4 g
I’m pretty sure the answer is D!
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
<em />
Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
<em />
The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
<em />
The phase's composition is as follows: 27
What is the alloy's composition?
Both have mass fractions of w a = w b = 0.5.
A-B alloy composition; C o = 57 wt% B - 43 wt% A
C b = 87 wt% B - 13 wt% A is the new phase composition.
Using the Lever rule, we can calculate the mole fraction (x i) or mass fraction (w i) of each phase of a binary equilibrium phase as follows:
W a = W b = 0.5
This provides us with;
0.5 = (C b - C o)/(C b - C a)
(87 - 57)/(87 - C a) = 0.5
30/0.5 = 87 - C a
60 = 87 - C a
C a = 87 - 60
C a = 27
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