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Sever21 [200]
3 years ago
14

What are two key characteristics of a good scientific investigation

Chemistry
1 answer:
AfilCa [17]3 years ago
8 0
There are more than two! Let's mention some: including experiments and observations into the study, having a control condition, closely following the devised procedure, having a hypothesis, etc...
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Create a chart that compares physical and chemical properties. Give two examples for each type of property
Zinaida [17]

<em>Answer:</em>

<em>Chemical properties:</em>

Those properties which change the chemical nature of matter.

<em>Example:</em>  

  •       Heat of combustion
  •       Enthalpy of formation

<em>Physical properties:</em>

Those properties which do not change the chemical nature of matter.

<em>Example</em>

  • B.P
  • M.P
  • F.P

<em>Differences between chemical and physical properties:</em>

       Chemical properties                                       Physical properties

1. Observed after the change bringing           1. Observed with out being

the change                                                            change

2. These changes the molecules                    2. only change physical state

3. Chemical identity changes                          3.Chemical identity not changes

4. Structure of material changes                     4.Structure of material not change            

5. Chemical reaction is needed                       5. No need of Chemical reaction

6. depend on composition                           6. Does not depend on composition

3 0
3 years ago
4. Solve the following heat flow problem, being sure to show all your work (you may either type your
Viktor [21]

Answer:

0.70 J/g.°C

Explanation:

Step 1: Given data

  • Mass of graphite (m): 402 g
  • Heat absorbed (Q): 1136 J
  • Initial temperature: 26°C
  • Final temperature: 30 °C
  • Specific heat of graphite (c): ?

Step 2: Calculate the specific heat of graphite

We will use the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1136 J / 402 g × (30°C - 26°C)

c = 0.70 J/g.°C

5 0
3 years ago
Please help! (no links)
Rus_ich [418]
It's D because I know bcz I'm smart 8 times 8 is 60-nart
5 0
3 years ago
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04
evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0
3 years ago
Part iv. Is the neutralization reaction enthalpy favored?
Burka [1]

Yes, it is a special case of enthalpy of neutralization.  

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

5 0
3 years ago
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