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algol [13]
3 years ago
13

Help plz:))) I’ll mark u brainliest ASAP 10 points

Chemistry
1 answer:
statuscvo [17]3 years ago
8 0

Answer: I believe it’s the second one

(F)

Explanation:

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Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of fructose, C6H12O6(s)
ziro4ka [17]

Answer:

6 C(s) + 6_{H2}(g) + 3_{O2}(g)

Explanation:

The chemical equation of the reaction is as follows

According to the question, data provided in the question is shown below:

C_6H_12O_6

The C = Carbon and its standard state is Graphite C(s)

The H = Hydrogen and its standard state is H_2 (s)

The O = Oxygen and its standard state is O_2 (s)

Now the chemical equation for the reaction is

6 C(s) + 6_{H2}(g) + 3_{O2}(g)

5 0
4 years ago
How many moles of H2O were produced during the combustion of CH4
Goryan [66]
<span>Reaction of combustion ( CH4 ) :


</span><span>CH4 + 2 O2 = CO2 + 2 H2O 
</span><span>   </span>↓                                    ↓<span>
1 mole CH4 produced  2 moles of H2O

hope this helps!





</span>
5 0
3 years ago
A confused student was doing an isomer problem and listed the following six names as different structural isomers of C7H16. a. 1
klasskru [66]

In case of heptane (C7H16) the following structural isomers are possible

shown in figure

a. 1-sec-butylpropane : this is actually 3-methyl hexane

b. 4-methylhexane : this is actually 3-methylhexane

c. 2-ethylpentane : this is actually 3-methyl hexane

d. 1-ethyl-1-methylbutane: 3-methylhexane

e. 3-methylhexane: correct IUPAC

f. 4-ethylpentane: This is actually 3-methylhexane

Hence all represent single isomer

4 0
3 years ago
Read 2 more answers
What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?
lukranit [14]
1 mol --- 22,4 dm³
2,1 mol --- X
X = 2,1×22,4
X = 47,04 dm³ = 47,04 L

C)
3 0
3 years ago
Read 2 more answers
Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
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