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3 years ago

What are 2 examples of ethical research behaviour?

Chemistry

1 answer:

FromTheMoon [43]3 years ago

6 0

Answer:

Respect
Confidentiality

Explanation:

*Hope this helps*

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A 45.7 block of an unknown metal at 74.2 °C is placed in 72.9 g of water at 15.9 °C. If the

g100num [7]

Answer:

0.934J/g°C

Explanation:

Using Q = mc∆T

However, in this question;

(Q)water = -(Q)metal

(mc∆T)water = -(mc∆T)metal

According to the information provided in the question;

For water;

m = mass = 72.9g

c = specific capacity of water = 4.184 J/g°C

∆T = 22.9 - 15.9 = 7°C

For metal;

m = mass = 45.7g

c = specific capacity of water = ?

∆T = 22.9 - 72.9 = -50°C

(mc∆T)water = -(mc∆T)metal

(72.9 × 4.184 × 7) = -(45.7 × c × -50)

2135.0952 = -(-2285c)

2135.0952 = 2285c

c = 2135.0952/2285

c of metal = 0.934J/g°C

3 0

2 years ago

How was your day????????

Trava [24]

Answer:

I literally just woke up for school hehe

3 0

2 years ago

Read 2 more answers

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resista

lina2011 [118]

Answer:

The ice chunk will take 2.47 seconds to hit the water.

Explanation:

Height from which ice chunk fell = h = 30.0 m

Initial velocity of the ice chunk = u = 0 m/s

Time taken by ice chunk to cover 30.0 m of height = t

Acceleration due to gravity = g $9.8m/s^2$

Using second equation of motion ;

$h=ut+\frac{1}{2}gt^2$

$30.0 m=0 m/s\times t+\frac{1}{2}\times 9.8 m/s^2\times t^2$

$t=\sqrt{\frac{30.0m/s\times 2}{9.8 m/s^2}}2.47 s$

The ice chunk will take 2.47 seconds to hit the water.

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2 years ago

Covers 70 precent of earths surface

Anastaziya [24]

Answer:

<h2>Water covers 70 percent of earth's surface.</h2>

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1 year ago

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

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3 years ago