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natima [27]
3 years ago
13

Which family in the periodic table has metals that are usually harder than other metals? Alkali metals Alkaline earth metals Lan

thanide series Transition metals
Chemistry
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

The correct answer is "Alkaline earth metals".

Explanation:

Alkaline earth metals are a group of elements that are located in group 2 of the periodic table. They are harder than alkaline metals, have brightness and are good electrical conductors. Hardness is the degree of scratch resistance a material offers.

Have a nice day!

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A liquid is placed in a closed container and time passes until
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Answer: Dynamic equilibrium

Explanation:

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You discover some bones that have roughly %s as much carbon-14 in them as
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Answer:

150 years i guess so? im nkt sure

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Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en
Bezzdna [24]

Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

  • 1.24 gallon * \frac{3.785L}{1gallon} *\frac{1000mL}{1L} = 4693.4 mL

We <u>calculate the mass of Octane</u>:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * \frac{8molCO_{2}}{1molOctane} * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * \frac{9molH_{2}O}{1molOctane} * 18 g/mol = 4.69 g H₂O

7 0
3 years ago
A balloon is inflated such that a radius is 7 cm what is the volume of air inside the balloon in milliliters
DedPeter [7]

Answer:

V=14373.33\ \text{mm}^3

Explanation:

It is given that,

The radius of the inflated balloon is 7 cm.

We need to find the volume of air inside the balloon in milliliters.

The balloon is in the shape of a sphere whose volume is given by :

V=\dfrac{4}{3}\pi r^3\\\\\text{Put r = 7 cm}\\\\V=\dfrac{4}{3}\times \dfrac{22}{7}\times 7\times 7\times 7\\\\V=1437.33\ \text{cm}^3\\\\\text{We know that, 1 cm = 10 mm}\\\\\text{So},\\\\V=14373.33\ \text{mm}^3

Hence, the volume of the air inside the balloon is 14373.33\ \text{mm}^3.

3 0
2 years ago
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
2 years ago
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