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Yanka [14]
2 years ago
15

The calorimeter pictured below shows the final temperature after calcium chloride was added to water. (The initial temperature w

as 25°C.) Which of the following best describes what occurred within the calorimeter?
A. Only Bond forming occurred

B. The bonds that formed were the most part stronger than the bonds that were broken

C. More energy was absorbed in breaking bonds than was released when new bonds form

D. As the powder dissolved, the breaking of bonds released energy
Chemistry
2 answers:
creativ13 [48]2 years ago
7 0

Answer:

The answer is acctualy B

Explanation:

I did the gizmo

max2010maxim [7]2 years ago
3 0
It’s probably D or c
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What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)
erik [133]

Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

Wavelength (λ) = 4.2×10¯⁷ m

Energy (E) =.?

Next, we shall determine the frequency of the wave. This can be obtained as follow:

Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

3×10⁸ = 4.2×10¯⁷ × f

Divide both side by 4.2×10¯⁷

f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

E is the energy.

h is the Planck's constant

f is the frequency

Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

E = 6.63×10¯³⁴ × 7.14×10¹⁴

E = 4.74×10¯¹⁹ J

Therefore, the energy of the wave is 4.74×10¯¹⁹ J.

5 0
3 years ago
PLS HELP QUICK ALOTTT OF POINTS
timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

0.7970773829 \ mol \ F

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

6 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

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3 years ago
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2 years ago
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