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alex41 [277]
1 year ago
8

Forming a hypothesis is accomplished through _______ reasoning.

Chemistry
2 answers:
vfiekz [6]1 year ago
4 0
Deductive reasoning involves making broad observations and examining possibilities to reach a specific, logical conclusion, which is required in the formation of a hypothesis. Conversely, inductive reasoning involves taking specific observations and making broad generalisations from these. Therefore the answer is A.

Hope this helps!
spayn [35]1 year ago
4 0

Answer:

a. deductive

Explanation:

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A crystal of graphite contains which type of bond?
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1 year ago
Read 2 more answers
What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the
bulgar [2K]
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>

4 0
2 years ago
Read 2 more answers
A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
Leona [35]

Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
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