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laiz [17]
3 years ago
5

PLS HELP!!! WILL GIVE BRAINLIEST!!! Really easy, I am just horrible at this stuff.

Chemistry
2 answers:
otez555 [7]3 years ago
6 0

Answer:

B: Igneous, C: Metamorphic, D: Sedimentary

Explanation:

Sorry, i dont know A

Fynjy0 [20]3 years ago
3 0
I think it’s c sorry if I’m wrong
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650.J is the same amount of energy as? <br>2720cal<br>1550cal<br>650.cal<br>2.72cal
Oksi-84 [34.3K]

<u>Ans: 650 J = 155 calories</u>


<u>Given:</u>

Energy in joules = 650 J

<u>To determine:</u>

The energy in calories

<u>Explanation:</u>

1 joule = 0.2388 calories

Therefore, 650 joules = 0.2388 calories * 650 J/1 J = 155 calories

4 0
4 years ago
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1.- ¿Cuál es la densidad del corcho, si un cubo que mide 1.50 cm de lado tiene la masa de 1.00 g?
Brrunno [24]

Answer: 2.50

Explanation:

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What is the pressure exerted by a force of 15.0 N over an area of 0.00372 m^2 ?
aleksandrvk [35]

Explanation:

Pressure = Force/Area

therefore,

P = 15/0.00372

= 4.03 X10^³N/m²

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2 years ago
(b) in the direct hydration method, ethene reacts with steam. The equation for this reaction is C2H4 + H2O = C2H5OH
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20 degree and 4000000N

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3 years ago
What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?
blondinia [14]

Answer:

The freezing point of the solution is -1.4°C

Explanation:

Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>

ΔT = kf×m×i

<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>

Molality of the solution is:

-moles CaBr₂ (Molar mass:

189.9g ₓ (1mol / 199.89g) = 0.95 moles

Molality is:

0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>

Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:

CaBr₂(s) → Ca²⁺ + 2Br⁻

3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.

Replacing:

ΔT = kf×m×i

ΔT = 1.86°C/m×0.253m×3

ΔT = 1.4°C

The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,

<h3>The freezing point of the solution is -1.4°C</h3>

<em />

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3 years ago
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