<span>Step 1 is to determine the mass of each part
Mass of Ca is 40.08 g
Mass of C is 12.01 g
Mass of O is 16.00 x 3 = 48.00 g
Step 2 is to determine the total mass of the compound
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g
Step 3 is to determine the % of each part using the following formula:
Mass of part / total mass x 100 =
40.08 / 100.09 x 100 = 40.04 % Ca
12.01 / 100.09 x 100 = 12.00 % C
48.00 / 100.09 x 100 = 47.96 % O
Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :)
40.04
+12.00
+47.96
=100.00</span><span>
</span>
Answer: -2m/s2
Explanation:
Using the following equation ; acceleration = Change in velocity / time
i.e a = v - u / t
where 'a' = acceleration
v = final velocity
u = initial velocity
t = time
Therefore; from the graph we have acceleration to be, 0 - 6m/s / 3s = -2m/s2
Answer:
C = (5/9) F - (160/9)
They both read equal at Z = - 40
Explanation:
We are looking for a linear function so we can write the following condition
Y = aX + b
Applying it to the exercise we got C = a F + b
Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212
0 = 32 a + b (1)
100 = 212 a + b (2)
From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)
and b = - (5/9)32 = - 160/9
Finally the linear function is C = (5/9) F - (160/9)
Both readings are equal at a Z number so
Z = (5/9) Z - 160/9
(4/9) Z = -160/9 and Z = - 40
Answer:
Convex mirrors are more study than flat mirrors.
Explanation:
This is the correct answer because i have already studied this
Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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