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lidiya [134]
3 years ago
13

Come People join meet this is the code to join meetpjv-ngby-iir​

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

what the pretada if u ask to many question like that u will get banned lol

Explanation:

You might be interested in
10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what
igor_vitrenko [27]

Answer:

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

3 0
3 years ago
Why can’t an atom lose or gain a proton
Veronika [31]

Explanation:

Atoms never gain protons; they become positively charge only by losing electrons. A positive ion is called a cation (pronounced: CAT-eye-on). You may have notice that the number of neutrons in each of these ions was not specified.

6 0
3 years ago
Read 2 more answers
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the ca
schepotkina [342]

Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

                                      c: specific heat of water = 4.186 J/gºC

                                     ΔT : change in temperature = 2.06 ºC

so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter  = C x  ΔT  where C: heat capacity of calorimeter = 69.6 ºC

                                     ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J +  143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

7 0
3 years ago
5.25 ml of substance A has a mass of 3.9 g and 6.24 ml of substance B has a mass of 4.4 g. Which liquid is more dense?
Viktor [21]
Substance A because it weighs less
5 0
3 years ago
What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




5 0
3 years ago
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