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3 years ago

12

Do physical changes transfer energy to and from their surroundings, yes or no?

1 answer:

GenaCL600 [577]3 years ago

5 0

I want to say the answer is yes.

You might be interested in

A. Energy is transferred to different forms

Luden [163]

Answer:

A. Energy is transferred to different forms .

Explanation:

Hello!

In this case, we need to consider the law of conservation of mass and energy which states that mass and energy cannot be neither created nor destroyed, just modified; it means we can rule out B. and C. so far.

Moreover, since D. is actually true for combustion reactions because they are used to provide energy in industrial operations, this is not the concern here because a combustion reaction is not considered.

Therefore the correct option is A. Energy is transferred to different forms as the energy provided by Rose is transferred to the pendulum system .

Best regards!

3 0

2 years ago

Write each mixed number as as a decimal. Use bar notation if the decimal is repeating.

aniked [119]

Number 1 is 4.875 Number 2 is 3.47 repeating

6 0

2 years ago

How old is the earth in millions of years

Paul [167]

A million years old
Duhhhh

6 0

2 years ago

Read 2 more answers

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

2 years ago

When light strikes chlorophyll molecules, they lose electrons, which are ultimately replaced by other electrons obtained by ____

fomenos

The answer is by "splitting water".

hope this helps

5 0

3 years ago