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2 years ago

Where else do you see Force & Motion in real-world experiences?

Chemistry

2 answers:

qwelly [4]2 years ago

7 0

Answer:

<h2>Daily traffic. </h2><h2>Amusement parks.</h2><h2>Gymnastic.</h2>

Explanation:

Force and Motion are everywhere like daily traffic, amusement parks and gymnastic. This to physical magnitudes are everywhere because all objects are relatively moving, and to do that a force is needed to being applied, because force means to give acceleration to a body, which can cause movement, in case it doesn't, it's said that the system is in equilibrium.

Paladinen [302]2 years ago

4 0

Answer:

a boy pulling a toy train, the boy is

interacting with an object while applying a force to it,

another example of contact forces is friction.

Explanation:

A contact force is when two interacting

objects

(btw love your pfp)

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A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal

devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0

2 years ago

How does one calculate the pH of a solution? O A. The pH is the product of [H+] and [OH-]. O B. The pH is the negative log of [O

blagie [28]

Answer:

I believe it is A

Explanation:

pH = -log[H+] and pOH = -log[OH−].

8 0

2 years ago

Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen

andrew-mc [135]

The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g

7 0

3 years ago

a) Provide equation of K of this reaction, use symbol " ^ " for exponents. That means 1000 = 10^3 and 1/100 is 10^(-2). b) How m

stealth61 [152]

Answer and Explanation:

a. The equation of K of this reaction is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

$K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}$

b. The moles of compound F is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

2 moles

Now, the mole of produced is

$= \frac{1}{4} \times \ moles\ of\ c$

Now, we will the value of c by using the above equation

$= \frac{1}{4} \times 2$

After solving the above equation we will get

0.5 moles

5 0

3 years ago

Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

Inessa [10]

Answer:

Gd(g) → $Gd^{+}(g) + e^{-}$ , $Gd^{+}(g)$ → $Gd^{2+}(g) + e^{-}$ , $Gd^{2+}(g)$ → $Gd^{3+}(g) + e^{-}$

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ $Gd^{+}(g) + e^{-}$

The second ionization energy is the energy required to remove next/second electron from $x^{+}$ ion. The second ionization energy is always higher than the first:

$Gd^{+}$ (g) → $Gd^{2+}(g) + e^{-}$

The third ionization energy is the energy required to remove third electron from $x^{2+}$ ion:

$Gd^{2+}$ (g) → $Gd^{3+}(g) + e^{-}$

5 0

2 years ago