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pshichka [43]
3 years ago
6

The degree or intensity of heat present and how we measure heat is called:

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
5 0
Ans: the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch. Thermal energy is the internal energy of an object due to the kinetic energy of its atoms and/or molecules.
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Describe in your own words, in terms of particle movement and energy, what occurs when a liquid is heated to its boiling point.
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D. The sun has less gravitational pull on Pluto because it is farther away.

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Removing seed casings from grains is SEPARATING. a soda bubble bubbling when it is opened is MIXING. a bright copper statue turning green is MIXING. remove salt from seawater is SEPARATING. water decomposing is SEPARATING.
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3 years ago
Which solution contains the smallest number of moles of sucrose (c12h22o11, molar mass = 342.30 g/mol)? 2,000 ml of a 5.0 × 10–5
Iteru [2.4K]

> 2,000 mL of a 5.0 × 10–5% (w/v) sucrose solution 

5.0 × 10–3 g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol

<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>

5 grams / 1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol

 <span>
> 20 mL of a 5.0 M sucrose solution </span>

5.0 M * 0.020 L = 0.1 mol

 

 

Answer:

<span>2,000 mL of a 5.0 ppm sucrose solution</span>

5 0
4 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
4 years ago
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