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sesenic [268]
3 years ago
8

Student follow up

Chemistry
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

Magnesium and calcium belong to the second group i. e. alkaline earth metals. They are known as earth metals because they are extracted from the earth. They are very reactive elements. Their reactivity increases when we go from top to bottom because the outermost electrons goes farther from the nucleus i. e. atomic radius increases so less energy is needed for its removal.

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If the density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cub
Tasya [4]

Answer:

1.35 × 10⁴ kg/m³ at 22 °C; 1.34 × 10⁴ kg/m³ at 100 °C

Explanation:

The cubic expansivity (γ) of a liquid is the fractional change in volume per unit change in temperature.

\gamma = (\frac{\Delta V }{ V_0} )(\frac{1 }{ \Delta T} )   Multiply by V₀ΔT and transpose

ΔV = γV₀ΔT  

and

V = V₀ + ΔV

===============

<em>At 0 °C </em>

Assume you have 1 m³ of Hg

ρ = m/V     Multiply by V and transpose

m = ρV

ρ = 1.36 × 10⁴ kg/m³

m = 1.36 × 10⁴ × 1 = 1.36 × 10⁴ kg

===============

<em>At 22 °C </em>

Assume that you have 1 m³ of Hg

γ = 180 × 10⁻⁶ K⁻¹

ΔT = 22 °C – 0 °C = 22 °C

ΔV = 180 × 10⁻⁶ × 22

ΔV = 3.96 × 10⁻³ m³      Calculate volume

V = 1 + 0.00396

V = 1.00396 m³             Calculate density

ρ = 1.36 × 10⁴/1.00396

ρ = 1.35 × 10⁴ kg/m³

===============

<em>At 100 °C </em>

ΔT = 100 °C – 0 °C = 100 °C

ΔV = 180 × 10⁻⁶ × 100

ΔV = 0.0180 m³      Calculate volume

V = 1 + 0.0180

V = 1.0180 m³          Calculate density

ρ = 1.36 × 10⁴/1.0180

ρ = 1.34 × 10⁴ kg/m³

4 0
4 years ago
Answer the following question.
aivan3 [116]

Answer:

B) The north and south Poles

Explanation:

when dealing with magnets they have a negative and Positive charge on their south and north poles. Negative attacks with positive and positive with negative so both poles on each magnet should react to each other

4 0
3 years ago
In the reaction ______H2+N2-&gt;2NH3 What coefficient should be placed in front of H2 to balance the reaction?
Paladinen [302]
3 should be placed in front of H2
5 0
3 years ago
What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I
mihalych1998 [28]

<u>Answer:</u> The energy released in the given nuclear reaction is 94.99 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}

We are given:

Mass of _{92}^{235}\textrm{U} = 235.043924 u

Mass of _{0}^{1}\textrm{n} = 1.008665 u

Mass of _{53}^{131}\textrm{I} = 130.9061246 u

Mass of _{39}^{89}\textrm{Y} = 88.9058483 u u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.1019761u)\times c^2

E=(0.1019761u)\times (931.5MeV)  (Conversion factor:  1u=931.5MeV/c^2  )

E=94.99MeV

Hence, the energy released in the given nuclear reaction is 94.99 MeV.

3 0
3 years ago
How was helium first discovered?
Igoryamba

Using spectral analysis of the Sun's emission spectra.

6 0
4 years ago
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