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Alla [95]
3 years ago
11

state whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting

of iron 4 electrolysis of water​
Chemistry
1 answer:
Artist 52 [7]3 years ago
5 0
1. rancidification fixation of water is CHEMICAL CHANGE

2. Tearing of paper is PHYSICAL CHANGE

3. Rusting if iron is CHEMICAL CHANGE

4. Electrolysis of water is CHEMICAL CHANGE
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Which of these formulas is the formula font the base calcium hydroxide
zhenek [66]
Ca(OH)2 is the answer for plato
4 0
4 years ago
What is one practical use of the paper battery
love history [14]

Answer:

A paper battery can work even if it is folded, cut or rolled up. A Paper battery consists mainly of carbon and paper; it can be used to power pacemakers within the body. A paper battery can be used both as a capacitor and battery. It is an ultra-thin storage device.Oct 10, 2014

Explanation:

5 0
3 years ago
A sample of sodium hydroxide (NaOH) has a mass of 160. 0 g. The molar mass of NaOH is 40. 00 g/mol. How many moles of NaOH does
Andrew [12]

4 moles of NaOH is present in the given 160 grams of NaOH.

<h3>How we calculate moles?</h3>

Moles of any substance will be calculated as:

n = W/M, where

W = given mass of NaOH = 160 grams (given)

M = molar mass of NaOH = 40 g/mole

Moles of NaOH is calculated as:

n = 160g / 40g/mol = 4 moles

Hence, moles of sodium hydroxide (NaOH) is 4 moles.

To know more about moles, visit the below link:

brainly.com/question/24322641

6 0
3 years ago
CaCO3(s) ∆→CaO(s) + CO2(g).
sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of CaCO₃ available:

\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}.

Molar mass of CO₂:

M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}.

Mass of the 0.400655 moles of \text{CO}_2 expected for the 40.1 grams of CaCO₃:

m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}.

What's the <em>percentage</em> yield of this reaction?

\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%.

7 0
3 years ago
Which of the following could be the rate equation for a first order reaction? please help this is timed
wel

Rate = k[A] = first order reaction

<h3>Further explanation</h3>

Given

Rate law

Required

A first-order reaction

Solution

The rate law : equation for the rate of chemical reaction

For reaction

aA + bB ⇒ C

The rate : r = k[A]ᵃ[B]ᵇ

The sum of exponents(a+b) is the reaction order

From the choice :

a. a+b = 2, second order reaction

b. a+b = 3, third order reaction

c. a+b = 4, fourth order reaction

d a+0 = 1, first order reaction

7 0
3 years ago
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