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3 years ago

At which temperature would a can of soda be most fizzy when it is opened?

2 answers:

7 0

Answer: at higher temperatures.
Justification:

1) Soda have CO₂ dissolved. Carbonation consists on that: dissolving CO₂ into water, leading to carbonated water.

2) The solution of a gas into a liquid is inversely related to the temperature: the lower the temperature the more gas gets dissolved.
So, in the manufacturing of soda, the CO₂ is added in cool water in a cool environment.

3) So, the higher the temperature after the soda is delivered, the more gas will be liberated when you open the can.

kiruha [24]3 years ago

4 0

The temperature would be 40c

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

Mazyrski [523]

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>

I hope it helps!

6 0

3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Scilla [17]

Answer:

The essence including its given problem is outlined in the following segment on the context..

Explanation:

The given values are:

Moles of CO₂,

x = 0.01962

Moles of water,

$\frac{y}{2} =0.01961$

$y=2\times 0.01961$

$=0.03922$

Compound's mass,

= 0.4647 g

Let the compound's formula will be:

$C_{x}H_{y}O_{z}$

Combustion's general equation will be:

⇒ $C_{x}H_{y}O_{z}+x+(\frac{y}{4}-\frac{z}{2}) O_{2}=xCO_{2}+\frac{y}{2H_{2}O}$

On putting the estimated values, we get

⇒ $12\times x=1\times y+16\times z=0.4647$

⇒ $12\times 0.01962+1\times 0.03922+16\times z=0.4647$

⇒ $0.27466+16z=0.4647$

⇒ $z=0.01187$

Now,

x : y : z = $0.01962:0.03922:0.01187$

= $\frac{0.01962}{0.0118}:\frac{0.03922}{0.0188}:\frac{0.0188}{0.0188}$

= $1.6:3.3:1.0$

= $3:6:2$

So that the empirical formula seems to be "C₃H₆O₂".

8 0

3 years ago

Which 5.0-milliliter sample of NH3 will take the shape of and completely fill a closed 100.0-milliliter container?(1) NH3(s) (3)

antoniya [11.8K]

NH3(g) will take the shape of and completely fill a closed 100.0 milliliter container.

5 0

3 years ago

A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop in 7.5 seconds? What is the cars acc

Marianna [84]

Answer:

The acceleration is: $4m/s^2$

Explanation:

Given

$u = 30.0m/s$ --- The initial velocity

$t = 7.5s$ --- time

$v = 0m/s$ -- The final velocity

Required

Determine the acceleration

To do this, we make use of the first equation of motion

$v = u - at$

We used negative because the car was coming to stop.

This gives:

$0 = 30 - 7.5 * a$

$0 = 30 - 7.5a$

Collect like terms

$7.5a = 30$

Solve for a

$a = \frac{30}{7.5}$

$a = 4m/s^2$

6 0

3 years ago

Estimate the solubility of m(oh)2 in a solution buffered at ph = 7.0, 10.0, and 14.0.

kherson [118]

<span>When M(OH)2 dissolves we have M(OH)2 which produces M2+ and 2OHâ’ pH + pOH=14 At ph =7; we have 7+pOH=14 pOH=14â’7 = 7 Then [OHâ’]=10^(â’pOH) [OH-] = 10^(-7) = 1* 10^(-7) At ph = 10. We have, pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4) Finally ph = 14. We have, pOH = 0 And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)... So [OH-] = 1</span>

4 0

3 years ago