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3 years ago

13

What would be the product when combustion reaction occurs in a small tightly closed container

2 answers:

Andreyy893 years ago

3 0

I would think Carbon monoxide and water.

Jet001 [13]3 years ago

3 0

Answer:

Incomplete combustion produces carbon monoxide along with some carbon diozide. Water is always produced during combustion. If the fuel is sulfur containing, carbod disulfide will result as well.

Explanation:

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What are the five states of matter?

Savatey [412]

<span>Solids, Liquids, Gases, Plasma, and Bose-Einstein Condensates. The main differences between these states of matter are the densities of the particles.</span>

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3 years ago

Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)

Marrrta [24]

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

3(46) kg-------------- 2(63) kg

x --------------- 7600000 kg

x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30

2 NO(g) + O2(g)---2 NO2(g)

2(30) ------------------2(46)

x ---------------- 8323809.5 kg

x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

4(17) -------------------- 4(30)

x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

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3 years ago

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation:

Leni [432]

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2 years ago

Highly liquid lava forms wide, shield-like mountains. True False

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2 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago