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Firlakuza [10]
3 years ago
13

This is the equation for the formation of magnesium chloride. Mg(s) + 2HCl(l) → MgCl2(aq) + H2(g) Which are the reactants and th

e products in this reaction? A. The reactants are MgCl2 (magnesium chloride) and HCl (hydrochloric acid); the products are Mg (magnesium) and H2 (hydrogen). B. The reactants are MgCl2 (magnesium chloride) and H2 (hydrogen); the products are Mg (magnesium) and HCl (hydrochloric acid). C. The reactants are Mg (magnesium) and H2 (hydrogen); the products are MgCl2 (magnesium chloride) and HCl (hydrochloric acid). D. The reactants are Mg (magnesium) and HCl (hydrochloric acid); the products are MgCl2 (magnesium chloride)and H2 (hydrogen).
Physics
1 answer:
iragen [17]3 years ago
5 0
The solution would be D. The reactants are Mg - magnesium and HCl - hydrochloric acid, as they appear on the left side of the equation and what are the starting substance for he reaction. The products of MgCl2 - magnesium chloride and H2 - Hydrogen gas. Are the resulting products from the rearranged atoms of the same elements.
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3 years ago
A clarinetist, setting out for a performance, grabs his 3.070 kg clarinet case (including the clarinet) from the top of the pian
Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

Apply Newton's second law of motion;

the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2

Therefore, the vertical acceleration of the case is 1.46 m/s²

4 0
3 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

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a = acceleration

Using the equation

v² = v₀² + 2 a d

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a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

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