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Sidana [21]
3 years ago
5

Consider the reaction below Na2CO3 (aq) + CaCl2 (aq) CaCO3 (s) + 2NaCl (aq) If the releases 39.4 kJ of energy, how many kilocalo

ries does it release? (1 cal = 4.184 J) (Round off answer to 2 decimal place)​
Physics
1 answer:
denis23 [38]3 years ago
3 0

Answer: 9.42

Explanation: yes

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I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)
Helga [31]
Hello! I can help you with this! 

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
4 0
3 years ago
Please help on this one?
qwelly [4]

Answer:

D. An image that is smaller than the object and is behind the mirror

5 0
4 years ago
Describe characteristics and identify oxidation-reduction(redox) reactions. <br><br> plz and thx
mars1129 [50]

Answer:

I hope This helps

Explanation:

The majority of oxidation-reduction (redox) reactions share two critical features. One is that a reduction happens in both, while equivalent oxidation occurs; they are combined.

8 0
3 years ago
In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the s
eimsori [14]

Question:

(a) What change in the comet’s velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet’s velocity? Would this change be noticeable? (The mass of the earth is 5.97×1024kg.)

Answer:

The answers to the question are;

(a) The change in the comet’s velocity produced by the collision is 2.86 × 10⁻⁶km/h or 7.944 × 10⁻⁷ m/s

(b) It would change our planet’s velocity by  6.70× 10⁻⁸ km/h or 1.86× 10⁻⁸ m/s

Change is too small to be noticeable

Explanation:

We not that the question is about conservation of liner momentum

Therefore we have, by listing out the known parameters

m₁ = Mass of "Deep Impact" = 372 kg

m₂ = Mass of Tempel 1 comet  = (0.1 to 2.5) × 10¹⁴ kg,

v₁ = Vaelocity of "Deep Impact" = 37000 km/h

v₂ = Velocity of Tempel 1 comet = 40000 km/h

From the principle of linear momentum, we have, for both bodies moving in opposite direction;

m₁×v₁ + m₂×v₂ = m₁×v₃ + m₂×v₃ since it was a crash, it is assumed that they both have the same final velocity

This gives

372 kg ×37000 km/h  - 0.1 × 10¹⁴ kg × 40000 km/h = (372 kg + 0.1 × 10¹⁴ kg )×v₃

13764000 kg·km/h - 4.0 × 10¹⁷  kg·km/h = 10000000000372×v₃

v₃ = (-399999999986236000 kg·km/h)/10000000000372 = -39999.999997 km/h ≈ - 40000  km/h in the direction of Deep Impact

Change in comet velocity

= 40000  km/h - 39999.999997 km/h

= 2.86 × 10⁻⁶km/h = 7.944 × 10⁻⁷ m/s

(b) If the colission is with earth, we have

m₃ = Mass of earth

From the principle of conservation of linear momentum, we have

m₂v₂+m₃ v₃ = (m₂ + m₃) v₄

v₃ = Initial velocity of Earth = 0 km/h

m₃ = Mass of Earth = 5.97 × 10²⁴ kg

Therefore, pluggin in the vaalues gives

0.1 × 10¹⁴ kg × 40000 km/h + 5.97 × 10²⁴ kg × 0 km/h = (0.1 × 10¹⁴ kg + 5.97 × 10²⁴ kg) × v₄

Therefore,

v₄ = (4.0 × 10¹⁷  kg·km/h + 0 kg·km/h)/ (5970000000010000000000000 kg)

= 6.70× 10⁻⁸ km/h = 1.86× 10⁻⁸ m/s

Change is too small

5 0
4 years ago
The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
Alchen [17]

1. Single dental x-rays: 5.0\cdot 10^{-11}m

The energy of the photon is

E=25 keV = 25,000 eV

Using the conversion factor

1 eV=1.6\cdot 10^{-19} J

we can convert it into Joules:

E=(25,000 eV)(1.6\cdot 10^{-19}J/eV)=4\cdot 10^{-15} J

The relationship between photon energy and wavelength is

\lambda=\frac{hc}{E}

where

h=6.63\cdot 10^{-34} Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

E is the energy

Substituting into the formula, we find

\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15} J}=5.0\cdot 10^{-11}m

2. Microtomography: 2.0\cdot 10^{-11} m

The energy of these photons is 2.5 times greater, so

E=(2.5)(4\cdot 10^{-15} J)=1\cdot 10^{-14} J

And by applying the same formula used at point 1, we find the corresponding wavelength:

\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14} J}=2.0\cdot 10^{-11}m

6 0
3 years ago
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