Pure substance can be __ and _

How do you label the delta E, on an energy diagram

vladimir1956 [14]

Delta energy on labelled diagram is attached below

4 0

3 years ago

How to balance _h2s+ _o2 = _h2o+ _s

goldfiish [28.3K]

Answer:

2H₂S + 1O₂ → 2H₂O + 2S

Explanation:

SOLUTION :-

Balance it by using 'hit & trial' method , and you'll get the answer :-

2H₂S + 1O₂ → 2H₂O + 2S

VERIFICATION :-

In reactant side of equation :-

Number of atoms in H = 2×2 = 4
Number of atoms in S = 2×1 = 2
Number of atoms in O = 1×2 = 2

In product side of equation :-

Number of atoms in H = 2×2 = 4
Number of atoms in O = 2×1 = 2
Number of atoms in S = 2×1 = 2

Number of atoms of each element is equal in both reactant & product side of equation. Hence , the equation is balanced.

5 0

3 years ago

3.Using Balanced Equations: C5H12+802–5CO2+6H20

Inessa [10]

Answer:

i think b is the answer hope this helps:))

8 0

3 years ago

Production of Jam from Crushed Fruit in Two Stages. In a process producing jam (C1), crushed fruit containing 14 wt % soluble so

Andrej [43]

Answer:

The total kg from the mixer: $m=2222.5 kg$

Evaporated water: $m_{water ev.}=791.1 kg$

Jam produced: $m_{jam}=1431.4 kg$

Explanation:

Step by step:

1) Kg of mixture from the mixer

It says that crushed fruit is added to the mixer with the sugar and pectin.

The fruit added is 1000 kg

The sugar added is 1.22 kg sugar/1 kg crushed fruit:

$m_{sugar}=1000 kg fruit * \frac{1.22 kg sugar}{1 kg fruit}=1220 kg sugar$

The pectin added is 0.0025 kg pectin/1 kg crushed fruit:

$m_{pectin}=1000 kg fruit * \frac{0.0025 kg pectin}{1 kg fruit}=2.5 kg pectin$

The total kg from the mixer:

$m=1000 kg fruit + 2.5 kg pectin + 1220 kg sugar=2222.5 kg$

2) Evaporated water

To calculate the evaporated water it's important to have in mind that the mix goes from a concentration of 14 wt% to 67 wt%. This difference is because of the evaporated water. So:

Initial: 1000 kg of fruit with 14 wt% of solids

$m_{solids}=1000 kg *0.14=140 kg$

This amount of solids is constant

Final: The mass of solids is the same but now it represents the 67 wt%

$m_{final}=140 kg *\frac{100}{67}=208.9 kg$

and

$m_{water ev.}=1000kg-m_{final}$

$m_{water ev.}=1000kg-208.9kg=791.1 kg$

3) Jam produced

$m_{jam}=m_{final}+m_{pectin}+m_{sugar}=208.9 kg + 1220 kg + 2.5 kg$

$m_{jam}=1431.4 kg$

6 0

3 years ago

How many molecules are in 13.2 g NO2?

DochEvi [55]

Answer:

No. of molecules= mass/molar mass ×avogadro no.

= 13.2/50×6.02×10°23

No. of molecules =1.59×10°23

Explanation:

5 0

3 years ago

Pure substance can be __ and __

Pure substance can be and