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Shtirlitz [24]
3 years ago
15

What type of energy transformation is taking place when natural gas is used to heat water

Chemistry
1 answer:
kifflom [539]3 years ago
8 0
That is thermal energy 
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Cyclic compound A has molecular formula C5H10 and undergoes monochlorination to yield exactly three different constitutional iso
RideAnS [48]

Answer:

A: 1,2-dimethylcyclopropane

Explanation:

The possible cyclic structure with formula C₅H₁₀ are shown in the image.

A is a cyclic compound. On monochlorination, A yields 3 products.

To have 3 products on monochlorination, there should be three different carbon atoms.

Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.

Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.

Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.

Considering structure 5, there are three different carbon atoms,  thus three different structure are formed and thus the A is structure 5.

4 0
3 years ago
Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um
Andre45 [30]

Answer:

\large \boxed{\text{0.0038 mol/L}}

Explanation:

1. Calculate the initial moles of acid and base

\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}

2. Calculate the moles remaining after the reaction

                   OH⁻     +     H₃O⁺ ⟶ 2H₂O

I/mol:      0.0053       0.005 00

C/mol:    -0.00500   -0.005 00

E/mol:      0.0003              0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}

8 0
3 years ago
2
timofeeve [1]

Answer:

B. The Shell

Explanation:

The shell covers and surrounds whats inside. It protects it!

4 0
3 years ago
Read 2 more answers
‼️Stoichiometry / moles problem: ‼️
BaLLatris [955]
Amount of NH3 = 6.39 mol

Ratio of NH3 : Ratio of product
4 : 10

Hence, amount of product formed = 6.39/4 * 10 = 15.975mol = 16.0mol (3sf)
6 0
3 years ago
What volume of 2.5% (m/v) koh can be prepared from 125 ml of a 5.0% (m/v) koh solution?
disa [49]
     the  volume  of   2.5% m/v    koh  which  can  prepared  from  125  ml  of  a  5%  koh  solution  is  calculated  using  the  following  formula

 m1v1=  m2 v2

M1= 5/100=  0.05
v1=   125
m2=2.5/100=0.025
V2=?
v2=  m1v1/m2

=0.05  x125  /0.025=250  ml
8 0
3 years ago
Read 2 more answers
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