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3 years ago

What's the chemical equation for the carbon dioxide bubbles in a homemade lava lamp?

Chemistry

2 answers:

Goshia [24]3 years ago

8 0

Answer:

Explanation:

because 25 is a prime number meaning its very hot

julsineya [31]3 years ago

6 0

Answer:

C₆H₈O₇+ 3NaHCO₃ --› Na₃C₆H₅O₇ + 3CO2 + 3H₂O

Explanation:

The reaction occuring in lava lamp is acid base reaction.

When you drop tablet into water the citric acid reacts with sodium bicarbonate and forms water, a salt, and bubbles of carbon dioxide gas.

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

Goryan [66]

Answer: 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$

mass of $Fe_2(SO_4)_3$ = $moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g$

Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

3 0

3 years ago

Determine the pH of a 5x10^-4 M solution of Ca(OH)2

miss Akunina [59]

Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻.

pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!

5 0

3 years ago

How can equilibrium of a closed system chemical reaction be disturbed? Select all that apply.

charle [14.2K]

The correct answers are :

Changing the volume of the system.

Changing the temperature of the system.

Equilibrium will remain unaffected if the concentration of products and reactants are kept the same, and the temperature of the system is kept constant.

As the system is closed, we cannot add or remove products or reactants.

Change in temperature will shift the chemical equilibrium towards the reactant or product depending on whether the reaction is exothermic or endothermic.

Also change in volume will shift the chemical equilibrium of a chemical reaction if the reactants or products or both are gases.

8 0

3 years ago

Read 2 more answers

8. (01.01 LC)

Simora [160]

The answer is false!

5 0

2 years ago