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uysha [10]
2 years ago
7

Please help find out code?

Chemistry
1 answer:
Nat2105 [25]2 years ago
4 0
You might need to take more pictures so we can see all the equations clearly
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A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18
Aleks [24]

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f)  only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

    C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

    ocatne = 703 kg/m³

    hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

7 0
3 years ago
3<br> En la troposfera los gases se mantienen constante y en movimiento.<br> (1 Point)
dexar [7]

Answer:

56

Explanation:

5 0
3 years ago
One lap of an engineered hamster track measures 255 cm. To run 10. meters, how many laps should the hamster run? Report your ans
Maru [420]

Answer:

3.9 laps

Explanation:

Step 1: Given data

Length of the hamster track (L): 255 cm

Total distance to be run (D): 10 m

Step 2: Convert "D" to centimeters

We will use the relationship 1 m = 100 cm.

10 m × (100 cm/1 m) = 1.0 × 10³ cm

Step 3: Calculate the number of laps (n) that the hamster should run

We will use the following expression.

n = D/L

n = 1.0 × 10³ cm/255 cm

n = 3.9

4 0
3 years ago
How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
3 years ago
Aluminium sulfate hydrate al2(so4)3.xh2o contains 13.63% al by mass. calculate x, that is, the number of water molecules associa
Advocard [28]
MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]

54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3

>>> Al₂(SO₄)₃·3H₂O <<<<
:)
8 0
3 years ago
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