Most adult amphibians can obtain oxygen through?

What kind of chemical reaction is this

mezya [45]

Answer: from the lies this is synthesis reaction

Explanation:

It's also oxidation-reduction reaction. Li is oxidised and

H is reduced

7 0

3 years ago

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a piece of newly synthesized material of mass 25.0 g at 80.0C is placed in a calorimeter containing 100.0 g of water at 20.0C. I

serg [7]

Answer:

The correct answer is 1.194 J/g.ºC

Explanation:

The heat released by the material is absorbed by the water. We put a minus sign (-) for a released heat and a plus sign (+) for an absorbed heat.

We know the mass of the material (mass mat= 25.0 g) and the mass of water (mass H20= 100.0 g) and the specific heat capacity of water is known (Shw=4.18 J/g.ºC), so we can equal the heat released by the material and the heat absorbed by water y calculate the specific heat capacity of the material (Shm) as follows:

heat released by material = heat absorbed by water

-(mass material x Shm x ΔT)= mass water x Shw x ΔT

-(25.0 g x Shm x (24ºC - 80ºC)= 100.0 g x 4.18 J/g.ºC x (24ºC-20ºC)

25.0 g x Shm x (56ºC) = 100.0 g x 4.18 J/g.ºC x 4ºC

⇒Shm= (100.0 g x 4.18 J/g.ºC x 4ºC)/(25.0 g x 56ºC)

Shm= 1.194 J/g.ºC

5 0

3 years ago

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A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago

How many chlorine particles are in 4.5 moles in sig figs

Mariana [72]

Answer:

2.71×10²⁴ particles

Explanation:

Applying Avogadro's constant

Avogadro's constant states that, The number of specified particles contained in a mole of a substance is equal to 6.02×10²³.

From the question,

If one mole of chlorine contains a particle of 6.02×10²³,

Therefore,

4.5 moles of chlorine will contain a particle of (4.5× 6.02×10²³)

⇒ 27.09×10²³

Hence the number of particle contained in 4.5 moles of chlorine ≈ 2.71×10²⁴ particles

6 0

3 years ago

Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin

Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

CH₄ + 4Cl₂ --------→ CCl₄ + 4HCl

1 mol 4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of Cl₂ = 71 g/mol

mass of Cl₂

mass in grams = no. of moles x molar mass

mass of Cl₂ = 4 mol x 71 g/mol

mass of Cl₂ = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

mass in grams = no. of moles x molar mass

mass of CH₄= 1 mol x 16 g/mol

mass of CH₄ = 16 g

So,

284 g of Cl₂ combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂

Apply unity Formula

284 g of Cl₂ ≅ 16 g of methane ( CH₄ )

33.4 g of Cl₂ ≅ X g of methane ( CH₄ )

By cross multiplication

X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

5 0

3 years ago