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lions [1.4K]
3 years ago
13

How much heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C

Chemistry
1 answer:
Tom [10]3 years ago
3 0

1426.58 J  and 340.90 calories heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C.

Explanation:

Data given:

mass = 28.4 gram

initial temperature = -23 degrees

final temperature = 1degress

change in temperature  ΔT = Tfinal - Tinitial

ΔT =  1 -(-23)

 ΔT = 24 degrees

specific heat capacity of ice cube c = 2.093 J/g C

Formula used:

q = mc ΔT

putting the values in the equation:

q= 28.4 x 2.093 x 24

  = 1426.58 J

ENERGY IN CALORIES:

340.90 calories is the energy is required in the process.

You might be interested in
If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M Na
bonufazy [111]

Answer:

A) 31.22

Explanation:

The reaction of sulfuric acid with NaOH is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.

<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>

0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =

3.7289x10⁻³moles H₂SO₄

And moles of NaOH that you require to neutralize the acid are:

3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =

7.4578x10⁻³ moles NaOH

Using a 0.2389M NaOH solution:

7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL

Right answer is:

<h3>A) 31.22 </h3>

5 0
3 years ago
We say that salts will dissociate but acids will react with water. We say that acids will
Paha777 [63]

Answer:

ionize

Explanation:

Acids are chemical substances that lose/donate their hydrogen ion (H+) when they react with water. This property of acids is termed IONIZATION. In a chemical reaction involving acids and bases, acids release their proton or hydrogen ion (H+) in the presence of water solutions to form a conjugate base, which is usually an anion.

For example, in the chemical reaction;

HX + H20 -------> X- + H30+

HX is the acid because it loses its electron to water and forms the anion, X-, which is the conjugate base. Hence, it can be said that acid HX ionizes in water.

7 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
3 years ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
Please please !!! help me I’ve been doing this for 2 hours
MatroZZZ [7]

Answer:

3.7 x 10^23

is the right answer

6 0
3 years ago
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