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poizon [28]
4 years ago
14

Imagine that you combine lemon juice with baking soda in a glass. You see gas bubbles forming along the sides of the glass. How

are the chemical properties of the gas different from the chemical properties of the baking soda and lemon juice?
Chemistry
2 answers:
olasank [31]4 years ago
6 0
<span>The gas that is formed is a different state of matter from the baking soda and lemon juice. The lemon juice has a different odor or smell than the gas. There is no color in the gas, while the lemon juice is yellow, and the baking soda is white.
^^^^^ hope this helps you! :) </span>
Sunny_sXe [5.5K]4 years ago
6 0

Answer:

Gas released in this process is a totally different chemical substance in comparison to the baking soda and lemon juice. It should be Carbon Dioxide formed during the reaction.

Explanation:

We have a baking soda in a glass: it is composed of water and baking soda (NaHCO3). We have lemon juice, its main component is the citric acid (H3C6H5O7). After that, both liquids are mixed and an acid-base reaction occurs.

The reaction releases gas, which should be carbon dioxide (CO2), according to this reaction:

H3C6H5O7 (aq) + 3 NaHCO3 (aq) → 3 CO2 (g) +3H2O (l) + Na3C6H5O7 (aq)

The products of this reaction are Water (H2O) a Salt and carbon dioxide. As carbon dioxide is in the gas phase, it will be released from the liquid mixture as bubbles.

Finally, we can conclude that Gas released in this process is a totally different chemical substance in comparison to the baking soda and lemon juice. It should be Carbon Dioxide formed during the reaction.  

You might be interested in
Which of the following represents the least number of molecues?
Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}

So, let´s get the number of molecules for each of the options.

a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules

b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules

c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules

d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules

d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules

the smalest number is in option a)

Best of luck.

7 0
3 years ago
Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.
padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

#SPJ4

5 0
1 year ago
Explain why science is a continuous profession of study.
xxMikexx [17]
Science is a continuous profession of study because new ideas are produced based on new evidence. Also, there are different topics of science such as climate change or new cures to different diseases.
3 0
4 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
4 years ago
ANYONE PLEASE HELP ME WITH MY CHEMISTRY HOMEWORK I REALLY NEED THE ANSWER RIGHT NOW I HOPE Y’ALL CAN HELP ME:(
nevsk [136]

Answer:

what grade is this ?

Explanation:

7 0
3 years ago
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