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grandymaker [24]
3 years ago
9

An investigation that is controlled

Chemistry
2 answers:
slega [8]3 years ago
5 0
A controlled variable 
NeTakaya3 years ago
3 0
Controlled variable
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Convert the following temperature to c 67k b) 1671k
kap26 [50]
  • Convert Kelvin to °c

(a) 67°c

Formula:-

k=°c+273

°c = 273 - k

°c = 273-67= 206 °c

(b) 1671 k

k=°c+273

°c = 273 - k

°c= 273-1671 = -1398°c

8 0
3 years ago
Critical practices to control the amount of solid waste include __________.
erica [24]
Reduce, reuse, recycle, compost. I think hope its helpful.
3 0
3 years ago
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A solution contains 14 grams of KCl in 100. grams of water at 40°C. What is the minimum amount of KCl that must be added to make
pashok25 [27]

Answer:

1386g

Explanation:

Solubility of a solute is the ability of the solute to mix into a liquid (the solvent). It measures the highest amount of substance mixed into a liquid solvent while they are both at equal amounts.

if 14g dissolved in 100cm3

  the amount that will dissolve in 1000cm3 = 14 x 1000/ 100

                                                                      =1400g

To form a saturated solution, 1400 - 14 = 1386g must be added.

8 0
3 years ago
Explain how the shorelines can affect the weather, in one to two sentences
defon

Answer:

Global warming causes sea-level rise as oceans expand, and makes storm patterns more energetic. Consequently it will affect most of the world's coastlines through inundation and increased erosion.

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>
5 0
3 years ago
Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air
Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

4 0
3 years ago
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