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SIZIF [17.4K]
2 years ago
6

Dos vehículos colisionan. Se sabe que antes del choque el vehículo 1 de masa "m", se desplazaba horizontalmente hacia la derecha

con U1=10 m/s y el vehículo 2 cuya masa es el 30 % de la masa del vehículo 1, se desplaza horizontalmente hacia la izquierda con velocidad de 8m/s. Si hay pérdida de energía en la colisión, la expresión matemática que mejor representa dicha situación es:
Chemistry
1 answer:
BabaBlast [244]2 years ago
8 0

Answer:

translate? ill edit this later

Explanation:

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Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:

<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2

As you can observe in the stoichiometric equations, 1 molecule of water can dissolve with 1 mole of urea.
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3 years ago
Choose all true statements about the water molecule: Choose all true statements about the water molecule: Bonding between the ox
amm1812

Answer: Water is a permanent electric dipole, having permanent charge separation.

Explanation:

Hydrogen bonding is an intermolecular force having partial ionic-covalent character.

In H_2O, O is a highly electronegative atom attached to a H atom through a covalent bond. The oxygen atoms being more electronegative gets partial negative charge and H atom gets partial positive charge. Thus water is permanent electric dipole.

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2 years ago
Jack lives in a small city in the Pacific Northwest, where the temperatures are cool all year and the sky is frequently cloudy a
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2 years ago
How does peer review affect research?
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4 0
2 years ago
Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen
alexgriva [62]

Answer:

V_{base}=24.04mL

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

2M_{base}V_{base}=M_{acid}V_{acid}

Therefore, for is to compute the volume of the used base, we proceed as shown below:

V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}

And we plug in to obtain:

V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL

Best regards!

8 0
3 years ago
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