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2 years ago

El teñido de una camiseta blanca con pintura es un cambio fisico o quimico

Chemistry

1 answer:

goblinko [34]2 years ago

4 0

respuesta: fisico

explicacion:Un cambio químico es cuando su estructura atómico es modificada en este caso lo único que estas haciendo es pintando la camiseta en ningún momento deja de ser camiseta

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Sara tells Michael she is 160 centimeters tall, while Michael says he is 60 inches tall. If there are 2.54 centimeters in an inc

makvit [3.9K]

Sara is taller. 160cm =160 cm while to convert inches to centimeters you just multiply. 60 x 2.54 = 152.4cm which is less than 160 so sara is taller.

4 0

2 years ago

Given the initial rate data for the reaction A + B –––> C, determine the rate expression for the reaction.

umka21 [38]

Answer:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

Explanation:

For the reaction A + B → C

The formula for rate of reaction is:

Δ[C]/Δt = k [A] [B]

As you have [A], [B] and Δ[C]/Δt information you can multiply [A] times [B] and take this value as X and Δ[C]/Δt as Y. The slope of this lineal regression will be k.

Thus, you must obtain:

y = 3,60x10⁻² X

Thus, rate of reaction is:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

I hope it helps!

3 0

2 years ago

Read 2 more answers

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

I hope it helps!

5 0

2 years ago

A cylinder contains 4.0 g of nitrogen gas. A piston compresses the gas to half its initial volume. Afterward, by what factor has

notka56 [123]

Answer:

The mass density will be doubled

Explanation:

Density is given by dividing the mass of a substance by its volume.
An increase in mass causes an increase in density and vice versa, while a decrease in volume causes an increase in density and volume.
Therefore, when the volume is halved, then the density will be doubled if the mass is kept constant.
This has no effect on the number of moles as the mass is constant.

8 0

2 years ago

How many molecules of glucose are in 1 l of a 100 mm glucose solution?

Karolina [17]

Answer is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of glucose solution.

V(glucose) = 1 L; volume of glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; amount of substance.

N(glucose) = n(glucose) · Na (Avogadro constant).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

6 0

2 years ago