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Tems11 [23]
3 years ago
9

How can you test a mineral for hardness? I NEED~ THIS NOW PLEASE

Chemistry
1 answer:
creativ13 [48]3 years ago
8 0
Courtney’s in minerals can vary due to impurities but is usually diagnostic we determine the relative hardness of minerals using a scale diversity bye mineralogists Freidrich Mohs The scale assigned hardness Tuten common index minerals and is based up on the ability of one mineral to scratch another
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Nitrous oxide (N2O) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is also a potent gr
scZoUnD [109]

Answer:

five half lives

Explanation:

Half-life is the time required for a quantity to reduce to half of its initial value.

How many half lives it would take to reach 3.13% form 100% of it's initial concentration:

100% - 50% : First Half life

50% - 25%: Second Half life

25% - 12.5%: Third Half life

12.5% - 6.25%: Fourth Half life

6.25% - 3.125%: Fifth Half life

This means it would take five half lives to get to 3.125% (≈ 3.13%) of it's original concentration.

3 0
2 years ago
Hydrogen produced from a hydrolysis reaction was collected over water and the following data was compiled.
Shkiper50 [21]

Answer:

  • 0.00358 mol

Explanation:

<u>1) Data:</u>

a) V = 93.90 ml

b) T = 28°C

c) P₁ = 744 mmHg

d) P₂ = 28.25 mmHg

d) n = ?

<u>2) Conversion of units</u>

a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter

b) T = 28°C = 28 + 273.15 K = 301.15 K

c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm

d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm

<u>3) Chemical principles and formulae</u>

a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.

b) Ideal gas equation: pV = nRT

<u>4) Solution:</u>

a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm

b) Moles of hygrogen gas:

pV = nRT ⇒ n = pV / (RT) =

n =  (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =

n = 0.00358 mol (which is rounded to 3 significant figures) ← answer

7 0
3 years ago
Read 2 more answers
ascorbic acid is a diprotic acid (Ka= 8.0x10^-5 and Ka2= 1.6x10^-12). What is the pH of a 0.260 M solution of ascorbic acid
zlopas [31]

The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.

<h3>How to calculate pH?</h3>

The pH of a solution can be calculated using the following expression:

pH = - log {H+}

According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;

pH = - log {0.260}

pH = 0.585

Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.

Learn more about pH at: brainly.com/question/15289741

#SPJ1

7 0
1 year ago
Draw the structure for 1-nitrobutane, making sure to add all non-zero formal charges.
TiliK225 [7]

The molecular structure of 1-nitrobutane is C_{4} H_{9} NO_{2}. The structure of 1-nitrobutane is shown below.

An atom's formal charge would be determined by the covalent model of chemical bonding, which assumes that almost all chemical bonds include equal sharing of electrons among all atoms, regardless their relative electronegativity.

The structure for 1-nitrobutane, making sure to add all non-zero formal charges

There are four kind of molecule present in 1-nitrobutane and they are carbon, hydrogen , nitrogen and oxygen. Nitrogen is bonded with two oxygen atom out of them one oxygen atom is attached with single bond and second oxygen atom is bonded with double bond. Nitrogen has positive charge whereas oxygen has negative charge.

It is a kind of alkane in with nitro group is attached with alkane group.

To know more about 1-nitrobutane

brainly.com/question/25045923

#SPJ4

7 0
1 year ago
What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi
Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL

3 0
3 years ago
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