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Charra [1.4K]
3 years ago
14

The volume of a cube that measures 10cm in length, 10 cm in width, and 10 cm in height is 1 L. True of False

Chemistry
2 answers:
algol133 years ago
8 0

Answer: False

Explanation:

1 Cubic centimeter (cm³) = 0.001 liter (L).

Volume of = 10 x 10 x 10 = 1000cm³

To convert cubic centimeters to liters, multiply the cubic centimeter value by 0.001 or divide by 1000.

barxatty [35]3 years ago
4 0

Answer:

volume of this cube = 1000 cc which is 1 L

so true

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A discus thrower throws a 1.6kg discus at 25m/s what's the kinetic energy?​
Ilya [14]

Answer:

<h2>500 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2} \\

where

m is the mass

v is the velocity

From the question

m = 1.6 kg

v = 25 m/s

We have

k =  \frac{1}{2}  \times 1.6 \times  {25}^{2}  \\  = 0.8 \times 625 \\  = 500

We have the final answer as

<h3>500 J </h3>

Hope this helps you

6 0
3 years ago
Which condition applies when water boils at 100oC
VLD [36.1K]
An ambient pressure of 1 atm
5 0
3 years ago
If the molar mass of the compound in problem 1 is 110 grams/mole, what is the molecular formula? With work?
tatyana61 [14]
C6H6O2 is the answer 
3 0
3 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
3 years ago
CaCO3(s)+2H*(aq) →Ca2+ (aq)+H2001+CO262)
Effectus [21]

Answer:CO2(g) will be formed at a faster rate in experiment 2 because more H+ particles can react per unit time

Explanation:

8 0
3 years ago
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