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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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A4 kg object is being pulled to the east by a 6 N force. What is the object's acceleration?

Dahasolnce [82]

1.5 ms⁻²

Explanation:

We understand that Force is also given as mass * acceleration;

F = Ma

If force is 6N and the mass is 4kg of the object, the a can be evaluated as follows;

6 = 4a

6/4 = a

1.5 = a

= 1.5m/s²

4 0

3 years ago

1 lb of CO2 occupies 0.6 ft^3 at a pressure of 200 psi. Determine the temperature of the system.

Anarel [89]

Answer: The temperature of the system is 273 K

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of carbon dioxide = 1 lb = 453.6 g (Conversion factor: 1 lb = 453.6 g)

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of carbon dioxide}=\frac{453.6g}{44g/mol}=10.31mol$

To calculate the temperature of gas, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide = 200 psia = 13.6 atm (Conversion factor: 1 psia = 0.068 atm)

V = Volume of carbon dioxide = $0.6ft^3=16.992L$ (Conversion factor: $1ft^3=28.32L$ )

n = number of moles of carbon dioxide = 10.31 mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the system = ?

Putting values in above equation, we get:

$13.6atm\times 16.992L=10.31mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=273K$

Hence, the temperature of the system is 273 K

8 0

3 years ago

The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati

Pepsi [2]

Answer: 3.72 M

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.66\times 10^{-3}s^{-1}$

t = age of sample = 15.0 minutes

a = let initial amount of the reactant = 10.0 M

a - x = amount left after decay process = ?

$15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}$

$\log\frac{100}{(a-x)}=1.43$

$\frac{100}{(a-x)}=26.9$

$(a-x)=3.72M$

The concentration of $H_2O_2$ in a solution after 15.0 minutes have passed is 3.72 M

7 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

How many miles of O2 can be produced by letting 12.00 miles of KC103 react?

Marat540 [252]

2KClO3 --> 2KClO2 + O2
12 6 (moles)
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2.

3 0

2 years ago