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BartSMP [9]
3 years ago
9

An air temperature of 95°C most often exists in which layer of the atmosphere?

Chemistry
1 answer:
Phoenix [80]3 years ago
4 0
Themosphere try that.
You might be interested in
The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.
Brrunno [24]

Answer:

The molecular formula of the compound is C_{8}H_{16}O_{4}. The molecular formula is obtained by the following expression shown below

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is  C_{2}H_{4}O

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = \left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}

n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Molecular formula = 4 \times C_{2}H_{4}O

Molecular formula is C_{8}H_{16}O_{4}

6 0
3 years ago
The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

3 0
4 years ago
8. Neil pogo sticks from his locker to his science class. He travels 8 m east then 8 m west
jeyben [28]

Answer:

12

Explanation:

5 0
3 years ago
Potassium permanganate, KMnO, and glycerin, C3H5(OH)3, react explosively according to the
Finger [1]

The volume of CO2 at STP =124.298 L

<h3>Further explanation</h3>

Given

Reaction

4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O

701,52 g of KMnO4

Required

volume of CO2 at STP

Solution

mol KMnO4 (MW=158,034 g/mol) :

mol = mass : MW

mol = 701.52 : 158.034

mol = 4.439

mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549

At STP 1 mol = 22.4 L, so for 5.549 moles :

=5.549 x 22.4

=124.298 L

4 0
3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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