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3 years ago

When a system performs work on the surroundings, the work is reported with a negative sign. True of False

1 answer:

5 0

The statement that is given above “When a system performs work on the surroundings, the work is reported with a negative sign” is true. When the surrounding performs work by the system, it is positive.

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Ted used a balance scale to weigh a 2.233 g sample of copper sulfate. Which of these measurements made by Ted is the most accura

Vika [28.1K]

There’s only one measurement the question doesn’t make sense

6 0

3 years ago

What do the elements Nickel and Technetium have in common?

nalin [4]

Answer:

nickel and Technetium are both solids

Explanation:

3 0

3 years ago

PH of 0,035M HCl? How to count if your calculator hasn't got -lg?

IrinaK [193]

Given information:

Concentration of HCl = 0.035 M

To determine:

pH of the solution

Explanation:

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

HCl → H+ + Cl-

Hence the concentration of H+ = Cl- = 0.035M

Now, pH measures the strength of H+ ions in a given solution. It is expressed as:

pH = -log[H+]

pH (HCl) = -log(0.035) = 1.46

Ans: pH of 0.035M HCl is 1.46

6 0

3 years ago

Did democritus belive atoms retain there identity as a chemical reaction

mihalych1998 [28]

No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .

4 0

3 years ago

Read 2 more answers

The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th

Fantom [35]

Answer: The amount remained after 151 seconds are 0.041 moles

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $4.82\times 10^{-3}s^{-1}$

t = time taken for decay process = 151 sec

$[A_o]$ = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}$

$[A]=0.041moles$

Hence, the amount remained after 151 seconds are 0.041 moles

7 0

4 years ago