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Anni [7]
2 years ago
13

¿que fuerza de debera aplicar en un embolo pequeño de una prensa hidraulica de 12cm2 de area ,para levantar un cuerpo de 70000N

de peso que se encuentra en el embolo mayor y cuya area es de 8.5 cm2
Chemistry
1 answer:
n200080 [17]2 years ago
7 0

Answer: un flechazo fácil

Explanation:

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The si unit for measuring acceleration is?
ehidna [41]

m/s² aka meter per second squared.

acceleration = change in velocity/time  

= distance/time

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time  

= m/s

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s

=m/s^2

7 0
3 years ago
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Mutations are more important to the evolution of a species because
EleoNora [17]

Answer: Mutations are important to the evolution of a species because is creates new DNA for a certain gene, creating a new allele.

Hope this helps!!! Good luck!!! ;)

8 0
3 years ago
Convert 18g of Li into atoms.
gizmo_the_mogwai [7]
The full decimal is 2.59328...
When you round up the answer is 2.6 atoms of Li
8 0
2 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka
raketka [301]

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

  • HA ⇄ H⁺ + A⁻
  • Ka = [H⁺][A⁻]/[HA]

We <u>calculate [H⁺] from the pH</u>:

  • pH = -log[H⁺]
  • [H⁺] = 10^{-pH}
  • [H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

  • [HA] * 0.66/100 = [H⁺]

We <u>calculate [HA]</u>:

  • [HA] = 0.138 M

Finally we <u>calculate the Ka</u>:

  • Ka = \frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]} = 6.02x10⁻⁶
3 0
2 years ago
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