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2 years ago

According to the chart of specific heats, which material requires the most heat to raise its temperature from 10°C to 30°C?

Chemistry

1 answer:

myrzilka [38]2 years ago

8 0

The material which requires the most heat to raise its temperature from 10°C to 30°C is oil.

<h3>What is the formula to calculate absorbed heat?</h3>

The formula which we used to calculate the amount of involved heat in relation with specific heat is:

Q = mcΔT, where

Q = absorbed heat
m = mass
c = specific heat
ΔT = change in temperature

Among the given materials, specific heat of oil is highest than other materials so will require maximum absorbed heat.

Hence, oil requires the most heat.

To know more about specific heat, visit the below link:

brainly.com/question/6198647

#SPJ1

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Jlenok [28]

To be honest, I can’t really see the question. So please next time just type it out lol

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2 years ago

A neutral atom in the ground state of Sulfur has its outer most valence electron in which orbital? f d

Natasha_Volkova [10]

<u>Answer:</u> The outermost valence electron enters the p orbital.

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in outer most orbital of an atom.

Sulfur is the 16th element of the periodic table having 16 electrons.

Electronic configuration of sulfur atom is $1s^22s^22p^63s^23p^4$

The number of valence electrons are 2 + 4 = 6

These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.

Hence, the outermost valence electron enters p orbital.

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3 years ago

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

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3 years ago

Hydrogen iodide is not produced by the same method as for hydrogen chloride.why??

lara [203]

Answer:

Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...

Using an ionic chloride and Phosphoric acid

H3PO4 + NaCl ==> HCl + NaH2PO4

H3PO4 + NaI ==> HI + NaH2PO4

H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.

2I-² === I2 + 2e-

Explanation:

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3 years ago

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Answer: Answer b

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2 years ago

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