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pogonyaev
2 years ago
12

I need help with this question please someone tell me the answer for this

Chemistry
1 answer:
Andrej [43]2 years ago
5 0
Aueyeowosy yo this is my answer

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At what temperature will aluminum have a resistivity that is three times the resistivity of tungsten at room temperature? (Assum
umka2103 [35]

Explanation:

At room temperature, resistivity of tungsten is 5.6 \times 10^{-8} and the resistivity of aluminium is 2.8 \times 10^{-8}.

Temperature coefficient of aluminium (\alpha) = 3.9 \times 10^{-3}

R_{a} = 3R_{c} = 5.1 \times 10^{-8} \ohm m

and,      R_{a} = R_{o}(1 + \alpha \times \Delta T)

                       = 2.82 \times 10^{-8} (1 + 3.9 \times 10^{-8} \times \Delta T)

                  \Delta T = 207.3^{o}C

                             T = (207.3 + 20)^{o}C

                                 = 227.3^{o}C

So, at 227.3^{o}C aluminium has resistivity which is three times the resistivity of tungsten at room temperature.

3 0
3 years ago
Explain how how radioactive decay can result in one element changing into another?
Marina CMI [18]

Answer:

bgyibvhxspijmkn

Explanation:

knshblubhobsuo

6 0
3 years ago
A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva
Sergeu [11.5K]

Answer:

  • <u>0.1255 M</u>

Explanation:

<u>1) Data:</u>

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

<u>2) Chemical reaction:</u>

The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:

  • Acid + Base → Salt + Water

  • H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

<u>3) Balanced chemical equation:</u>

  • H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

<u>4) Stoichiometric mole ratio:</u>

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

  • 1 mole H₂SO₄ : 2 mole NaOH

<u>5) Calculations:</u>

a) Molarity formula: M = n / V (in liter)

   ⇒ n = M × V

b) Nunber of moles of acid:

  • nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

  • nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

  • nₐ = nb

  • 0.1053 × (17.88 / 1,000) =  Mb × (15.00 / 1,000)

  • Mb = 0.1053 × 17.88  / 15.00 = 0.1255 mole/liter = 0.1255 M
3 0
3 years ago
Read 2 more answers
If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
2 years ago
If 0.251moles of H2O gas are produced, how many liters of oxygen gas was used?
Alborosie

Answer:

o.251 prduces 45.7L of oxogen

Explanation:

hope this helps

4 0
3 years ago
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