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3 years ago

ΚΕ: m.22 2 What is the kinetic energy of 10kg dog running with a velocity of 12m/s?

Chemistry

1 answer:

BigorU [14]3 years ago

3 0

Answer:

The answer is

<h2>720 Joules</h2>

Explanation:

The kinetic energy of a body can be found by using the formula

where

m is the mass

v is the velocity / speed

From the question

mass = 10 kg

velocity = 12 m/s

Substitute the values into the above formula and solve

That's

<h3> $KE = \frac{1}{2} \times 10 \times {12}^{2} \\ = 5 \times 144 \: \:$ </h3>

We have the final answer as

<h3>720 Joules</h3>

Hope this helps you

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Can you treat a sulfuric acid burn with sodium hydroxide

Alina [70]

Answer:

I think yes

Explanation:

sorry i havent done these type a questions in a while

3 0

2 years ago

What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm

disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to $\frac{5.71}{286}$ = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is $\frac{0.0199 X 1000}{250}$ = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³

3 0

3 years ago

The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof

sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O ⇄ H₃O⁺ + F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0

2 years ago

The majority of water molecules moving across plasma membranes by osmosis do so via a process that is most similar to ____.

ValentinkaMS [17]

Answer:

The answer is Facilitated Diffusion

Explanation:

Osmosis is most similar to facilitated diffusion.

Osmosis can be defined as : Osmosis is the movement of water molecules across a semi-permeable membrane, down a concentration gradient from a low concentrated solution into a more concentrated solution.

Facilitated diffusion is defined as: Facilitated diffusion is the flow of molecules requiring the aid of a protein, across a membrane, down a concentration gradient from a low concentrated solution into a more concentrated one.

4 0

3 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago