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Nata [24]
3 years ago
9

A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate

the wet density of the CL sample. (b) A small piece of the CL sample had a wet weight of 140.9 grams and a weight of 85.2 grams after drying. Compute the water content. (c) Compute the dry density of the CL sample
Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

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Nuetrik [128]

Answer:

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Rule 1. The element with the lower group number is written first in the name; the element with the higher group number is written second in the name. Exception: when the compound contains oxygen and a halogen, the name of the halogen is the first word in the name.

Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name.

Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the root of the element name (e.g., fluorine = F, "fluoride" = F-; sulfur = S, "sulfide" = S2-).

Rule 4. Greek prefixes are used to indicate the number of atoms of each element in the chemical formula for the compound. Exception: if the compound contains one atom of the element that is written first in the name, the prefix "mono-" is not used.

Explanation:

5 0
3 years ago
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Spell out the full name of the compound.
Talja [164]

Answer:

2–methyl–3–propylheptane.

Explanation:

The name of the compound can be obtained as follow:

1. Identify the functional group of the compound.

2. Determine the name of the compound by determining the longest continuous carbon chain.

3. Identify the substituent group attached.

4. Locate the position of the substituent group by giving it the lowest possible count. Where there are two or more different substitutent group attached to the compound, name them alphabetically.

5. Name the compound by combining

the above steps.

Thus, we can name the compound as follow:

1. The compound has only single bond. Therefore, the compound is an alkane.

2. The longest continuous carbon chain is 7. Therefore, the parent name of the compound is heptane.

3. The substitutent groups attached are:

Propyl i.e –C3H7

Methyl i.e –CH3

4. The substituent groups are located at:

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6 0
2 years ago
What is the molarity of a naoh solution if 39.1 ml of a 0.112 m h2so4 solution is required to neutralize a 25.0-ml sample of the
aleksandrvk [35]
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry  of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL  - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
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8 0
3 years ago
How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.
lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

4 0
3 years ago
So2+o2 = so3 método de tanteo​
gulaghasi [49]

Answer:

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Explanation:

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7 0
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