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dalvyx [7]
2 years ago
9

PLEASE HELPPP :[

Chemistry
1 answer:
kkurt [141]2 years ago
5 0

Answer:

C. The fruit and the hamburgers were affected by an increase in heat energy.

Explanation:

One claim Harvey can use to support the examples from his experiment is that the hamburgers and fruits were affected by an increase in the heat energy.

  • This chemical change is one that is solely driven.
  • This action Harvey is carrying out is cooking
  • When meals are cooked, the raw substances undergoes chemical change via the action of heat.
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Select the pair that consists of a base and its conjugate acid in that order. CO32−/CO22−
lana [24]

Answer: The pair that consists of a base and its conjugate acid in that order.NH_3/NH_4^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

H_3PO_4\rightarrow H_PO_4{2^-}+2H^+

H_2CO_3\rightarrow HCO_3^-+H^+

NH_3+H^+\rightarrow NH_4^+

HCO_3^-\rightarrow CO_3^{2-}+H^+

NH_3  is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it formsNH_4^+  which is a conjugate acid.

3 0
3 years ago
Difference between qualitative and quantitative data in chemistry
const2013 [10]
Quantitative data is numerical.
Qualitative data is non-numerical.
Hope this helps.
have a great day.
7 0
3 years ago
An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molal
galina1969 [7]

Answer:

[HOCH₂CH₂OH] = 24.1 m

Explanation:

Ethylene glycol → HOCH₂CH₂OH

60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.

Solution mass = Solute mass + Solvent mass

100 g = 60 g + Solvent mass

Solvent mass = 40 g

Molality are the moles of solute contained in 1kg of solvent.

We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles

We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg

Molality → 0.967 mol / 0.04 kg = 24.1 m

4 0
4 years ago
Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen
andrew-mc [135]
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O 
the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
7 0
4 years ago
A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine
bagirrra123 [75]

Answer:

The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

Initial volume V_{1} = 200 ml = 0.0002 m^{3}

Initial temperature T_{1} = 296 K

Initial pressure P_{1} = 101.3 K pa

Final temperature T_{2} = 336 K

Final pressure P_{2} =  K pa

Relation between P , V & T is given by

P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }

Put all the values in the above equation we get

101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )

V_{2} = 0.000151 m^{3}

This is the final volume of the sample of gas.

4 0
3 years ago
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