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mojhsa [17]
3 years ago
13

An example of a scientific law is the law of conservation of mass, which states that matter is neither created nor destroyed in

a chemical reaction.
Based on what you know about scientific laws in general, classify the following statements regarding the law of conservation of mass as true or false.
A) It was developed from past observations
B) It is subject to experimentation and revision.
C) It explains why mass is conserved.
D) It predicts future observations
Chemistry
1 answer:
Amanda [17]3 years ago
4 0

Answer:A) It was developed from past observations- true

It is subject to experimentation and revision.- false

It explains why mass is conserved- false

It predicts future observations-true

Explanation:

Before a hypothesis is proclaimed to be a law in science, numerous observation must have confirmed its validity and rigorous experimentation under carefully controlled conditions usually precedes the acceptance of a hypothesis as a law. The law of conservation of mass was developed from numerous past observation that proved its validity.

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Please help me for question 1 and 2
likoan [24]

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

b)

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

4 0
3 years ago
What is a living thing that can carry out life process by itself
9966 [12]

Answer:

All Living things

Explanation:

All living things are capable of carrying out all life processes. Some examples of such organisms are plants, animals, microorganisms (such as bacteria)

And organism too

6 0
3 years ago
Read 2 more answers
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and
Lunna [17]

The pH of the unknown solution is 3.07.

<u>Explanation:</u>

<u>1.Find the cell potential as a function of pH</u>

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0   lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

<u>2. Find the Q  value</u>

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse)  × 298.15

K × 2pH /2×96 485  C⋅mol

( inverse)

E cell= 0.0592 V × pH

<u>3. Finding the pH value</u>

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0
3 years ago
2. Mitosis is great for repair and damage. TRUE OR FALSE
inn [45]

Answer:

TRUE

HOPE THIS HELPS!

Explanation:

3 0
3 years ago
Read 2 more answers
Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen
Oksanka [162]
<h3>Answer:</h3>

10.6 mol NO

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

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

[RxN] 5 mol O₂ → 4 mol NO

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 10.56 \ mol \ NO

<u>Step 4: Check</u>

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

10.56 mol NO ≈ 10.6 mol NO

5 0
2 years ago
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