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Darina [25.2K]
4 years ago
7

Make a quantitative and qualitative statement about the following:. water, carbon, iron, hydrogen gas, sucrose, table salt, merc

ury, gold, air.
Chemistry
2 answers:
gladu [14]4 years ago
8 0
The elements in this list are mercury, gold, iron, carbon and hydrogen. The compounds in this list, on the other hand, are sucrose, table salt, water and air. Elements are composed only of one substance while compounds are composed of two or more substances.
EastWind [94]4 years ago
4 0

Explanation:

Quantitative statement provide us information regarding the quantity which can be measured or can be written down in numeric form.

Qualitative statement provide us information regarding the quality or trat which cannot be measured.

1) Water has density of 1 g/ml.

Water is colorless in color.

2) Carbon has 6 electrons , 6 protons and  6 neutrons in it atom.

Carbon is an element which has black color and non metal.

3) Iron has boiling point of 2,862 °C.

Iron is a metal and good conductor of electricity.

4)Hydrogen gas has two hydrogen atom in its molecule

Hydrogen gas is a colorless gas.

5) Sucrose is an organic molecule made up of three different atoms

:carbon, oxygen, hydrogen.

Sucrose is termed as table sugar which tastes sweet to our tongue.

6) Table salt is compound made up of two atoms of sodium and chlorine.

Table salt is white in color.

7) Mercury has melting point of -38.83 °C

Mercury is liquid at room temperature.

8) Gold has an atomic mass of 198.97 g/mol.

Gold is shinny hard metal.

9) Air contains 78% of nitrogen , 21 % of oxygen and 1 % of other gases.

Air is a colorless mixture of gases which are the integral of the Earth planet.

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Hey there Abigaildonaimor,

A cyclist rode at an average speed of 10mph for 15 miles. How long was the ride?

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Time = Distance / Rate
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         = 1.5 hours

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6 0
4 years ago
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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
Which of the elements listed would have the greatest electronegativity?
disa [49]

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6 0
3 years ago
The specific heat of silver is 0.234 j/(g·c). calculate the amount of energy that is needed to raise the temperature of 175 g of
Elodia [21]
Heat energy is supplied to materials and can cause an increase in temperature of the material. the formula is as follows
H = mcΔt
where H - heat energy 
m - mass of material 
c - specific heat 
Δt - change in temperature  - 40.0 °C - 22.5 °C = 17.5 °C
substituting the values 
H = 175 g x 0.234 Jg⁻¹°C⁻¹ x 17.5 °C
H = 716.6 J 
716.6 J is required 
3 0
3 years ago
How much NaAlO2 (sodium aluminate) is required to produce 2.15 kg of Na3AlF6?
Harrizon [31]

Answer:

839 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r:              81.97                209.94

                   NaAlO₂ + … ⟶ Na₃AlF₆ + …

Mass/g:                                   2150

1. Calculate the <em>moles of Na₃AlF₆ </em>

Moles of Na₃AlF₆ = 2150 × 1/209.94       Do the operation

Moles of Na₃AlF₆ = 10.24 mol Na₃AlF₆

2. Calculate the <em>moles of NaAlO</em>₂

The molar ratio is 1 mol NaAlO₂:1 mol Na₃AlF₆

Moles of NaAlO₂ = 10.24 × 1/1 = 10.24 mol NaAlO₂

3. Calculate the <em>mass of NaAlO₂ </em>

Mass of NaAlO₂ = 10.24 × 81.97             Do the multiplication

Mass of NaAlO₂ = 839 g

5 0
4 years ago
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