Mass of K = 39.1g/mole
Mass of H = 1g/mole
Mass of C = 12g/mole
Mass of O = 16g/mole
Mass of KHCO3 = 39.1 + 1 + 12 + 3*16 = 100,1g/mole
100% ------------ 100.1g of KHCO3
x% ----------------- 12g of C
x = 11,99% of C in KHCO3
100% -------------- 100.1 of KHCO3
y% --------------------3*16g of O
y = 47,95% of O in KHCO3
100% ------------- 100.1g of KHCO3
z% ---------------- 1g of H
z = 0,99% of H in KHCO3
11,99% + 47,95% + 0,99% = 60,93% ≈ 61%
<u>answer: C</u>
Answer:
1.22 g/cm³
Explanation:
Density = mass ÷ volume
Mass = 10.04 grams
Volume = 8.21 cubic centimeters
Density = mass ÷ volume
= 10.04 ÷ 8.21
= 1.2228989037758 g/cm³
Approximately
1.22 g/cm³ to 2 significant figure.
The final answer should be approximated to 2 significant figure because the mass and volume of the objective given is in 2 significant figure
Answer:
pH = 9.25
Explanation:
Reaction of ammonia (NH₃) with hydrochloric acid (HCl) is:
NH₃ + HCl → NH₄⁺ + Cl⁻
As equal volumes of a 0.020M solution of NH₃ and a 0.010M solution of react, molarity of NH₃ after reaction is 0.010M and 0.010M of NH₄⁺.
For the buffer NH₄⁺/NH₃ H-H equation to find pH of solution is:
pH = pKa + log₁₀ [NH₃] / [NH₄⁺]
pH = 9.25 + log₁₀ [0.010M] / [0.010M]
<em>pH = 9.25</em>
Answer:
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Answer:
<em>Chemical</em><em> </em><em>properties</em><em> </em><em>of </em><em>each </em><em>elements </em><em>are </em><em>determined</em><em> </em><em>as </em><em>the </em><em>elements</em><em> </em><em>of </em><em>the </em><em>el</em><em>e</em><em>ctronic </em><em>conf</em><em>i</em><em>guration </em><em>and </em><em>particularly</em><em> </em><em>by </em><em>its </em><em>outermost </em><em>valence </em><em>electrons </em><em>in </em><em>addition</em><em> </em><em>the </em><em>total </em><em>number</em><em> </em><em>of </em><em>electron</em><em> </em><em>shells </em><em>an </em><em>atom </em><em>determines </em><em>to </em><em>which </em><em>per</em><em>i</em><em>od </em><em>it </em><em>belongs</em><em> </em><em>to </em><em>.</em>
<em><u>maybe </u></em><em><u>this </u></em><em><u>might </u></em><em><u>help </u></em><em><u>u</u></em>