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Anestetic [448]
2 years ago
11

Scientific Inquiry begins with?

Chemistry
1 answer:
netineya [11]2 years ago
4 0
The answer is observations
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Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
Which chemical symbols will complete the equation for this decomposition reaction?
kondaur [170]

Answer:

Decomposition of KI:

2KI → 2K + 2I

6 0
3 years ago
Read 2 more answers
What element has 4 valence electrons and 5 electron shells?
marta [7]

Answer:  Tin (Sn)

Explanation:  The electron configuration for tin (Sn) is shown in the picture.  It's last electrons are:

5s^2 4d^10 5p^2

The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.

8 0
3 years ago
_____ NaPO4+_____KOH
Artist 52 [7]
NaPO4 + KOH -> KPO4 + NaOH
already balance
6 0
3 years ago
The radioactive element​ carbon-14 has a​ half-life of 5750 years. a scientist determined that the bones from a mastodon had los
Monica [59]
Radioactive elements obey 1st order  kinetics,

For 1st order reaction, relation between rate constant (k) and half life [t(1/2)] is,
k = \frac{0.693}{t(1/2)} =  \frac{0.693}{5750} = 1.205 X 10^-^4 hr^-^1

Also, for 1st order reaction, we have
t = \frac{2.303}{k} log  \frac{\text{initial conc.}}{\text{final conc.}}
 
Given that: <span>the bones from a mastodon had lost 78.5​% of their​ C14,
</span>∴ initial conc. of C14 = 100%, conc. of C14 left after time 't' = 21.5%

∴t = \frac{2.303}{1.205 X 10^(-4)} log \frac{\text{100}}{\text{21.5}} = 1.2758 X 10^4 hours
5 0
3 years ago
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