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mylen [45]
2 years ago
15

What reaction/soccur between an acid and base.

Physics
2 answers:
Elena-2011 [213]2 years ago
6 0

Answer:

neutralization reaction

Explanation:

The reaction of an acid with a base is called a neutralization reaction.

quester [9]2 years ago
4 0

es una reacción química que ocurre entre un ácido y una base produciendo sal y agua. ... Se mezcla un ácido fuerte con una base fuerte: Cuando esto sucede, la especie que quedará en disolución será la que esté en mayor cantidad respecto de la otra.

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The fluid inside the hydraulic jack has a pressure of 30,000 Pa. If the surface of
aleksley [76]

Explanation:

p = F /A

F = P×A

F = 30,000 Pa / 0.1 m²

F = 300,000 N

5 0
2 years ago
Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and
Galina-37 [17]

Answer:

1.08x10⁻⁷

Explanation:

F=(GM₁M₂)/r²

=((6.67x10⁻¹¹)(70)(52))/(1.5²)

=2.42788x10⁻⁷/2.25

=1.07905778x10⁻⁷

≈1.08x10⁻⁷

3 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd
Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ)  β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

    = 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

    = 1.07 × 10³(1 - [0.024]²)

    =  1.07 × 10³(1 - 0.000576)

    = 1.07 × 10³(0.999424)

    = 1069.38 gallons

7 0
3 years ago
Which group number below represents the Oxygen Group?
worty [1.4K]

Oxygen has an atomic number 8, because it has 8 protons and 8 electrons.

The first shell of an atom can hold up to 2 electrons but oxygen has 8 electrons, in that eight electrons 2 are in the first shell, so it has 6 more electrons left. The second shell can hold up to 8 electrons, oxygen has only 6 more electrons after the first shell is full, so it will have 6 electrons in the second shell

From this we know that oxygen has 2 shells so it is in period 2, and by counting from left to right, the sixth box in period 2 lies on group 16

Therefore Oxygen lies on group 16 and period 2

Happy to help :)

If you need my help for any other question, feel free to ask

8 0
3 years ago
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