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3 years ago

What is the value of making an observtion during an investigation

Chemistry

1 answer:

prisoha [69]3 years ago

3 0

Answer:

it helps you gather imformation which makes it easier to crack a crime

Explanation:

i alr explained

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SO2+O2 SO3 how many moles of O2 needed to combust 100.0g of sulfur dioxide

Oksanka [162]

2SO2 + O2 ------> 2SO3

1) M(SO2)= 32.0 + 2*16.0 = 64 g/mol

2) 100.0 g SO2 * 1 mol SO2/64 g SO2 = 1.5625 mol SO2

3) 2SO2 + O2 ------> 2SO3

2 mol 1 mol

1.5625 mol x mol

x= 1.5625/2=0.78125 ≈ 0.7813 mol O2

Answer: 0.7813 mol O2.

6 0

3 years ago

The common Eastern mole, a mammal, typically has a mass of 83 g, which corresponds to about 8.3 moles of atoms. (A mole of atoms

Alchen [17]

Answer:

1.6605 × 10⁻²³ g / atom , 9.99 amu

Explanation:

Atoms present in 1 mole = 6.022 × 10²³ atoms

Given that

Moles of common Eastern mole = 8.3 moles

So,

Atoms in 8.3 moles are:

8.3 moles = 6.022 × 10²³ × 8.3 atoms

Also,

Average mass = Mass of the sample / Number of atoms

Mass of sample = 83 g

Average mass = 83 g / 6.022 × 10²³ × 8.3 atoms = 1.6605 × 10⁻²³ g / atom

In amu :

1 amu = 1.66 × 10⁻²⁴ g / atom

So,

Average mass = 1.6605 × 10⁻²³ / 1.66 × 10⁻²⁴ amu = 9.99 amu

6 0

3 years ago

Consider the single‑step, bimolecular reaction. CH 3 Br + NaOH ⟶ CH 3 OH + NaBr When the concentrations of CH 3 Br and NaOH are

saw5 [17]

Answer:

0.014 M/s

Explanation:

The rate of a chemical reaction is defined as the change in the concentration of the products or reactants over a specific period of time. The reaction rate can be calculated by dividing the change in concentration of the reactants products by the time elapsed for the given reaction. The reaction rate can also be calculated bu using:

$r = k*[CH_{3}Br]*[NaOH]$

where:

r is the rate of the reaction, k is the rate constant, and [] signifies the concentration of the compound.

When the concentrations of $CH_{3}Br$ and NaOH are both 0.140 M, the rate of the reaction is 0.0070 M/s.

Therefore, when the concentration of $CH_{3}Br$ is doubled, the rate of the reaction will also be doubled, i.e. r = 2* 0.0070 = 0.014 M/s

3 0

3 years ago

3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa

Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

7 0

3 years ago

To completely neutralize a 0.325 g sample of pure aspirin, 15.50 mL of a sodium hydroxide solution is added. If 16.25 mL of the

Kruka [31]

<span>a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+

b. Should form 5.11 X 10^-3 mol Cu(OH)2

c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc

d. 5.11 X 10^-3 mol Mg X 24.3 g/mol = 0.124 grams Mg</span>

3 0

3 years ago