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anzhelika [568]
3 years ago
13

f 37.4 grams of water decompose at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced?

Chemistry
1 answer:
Allisa [31]3 years ago
4 0
2H2O=2H2+O2
37.4g H2O(1 mol/18.02)=2.07547 mol H2O
PV=nRT
(1.30)(V)=(2.07547)(.0821)(297)
Vwater=38.92898L
38.92898L (1 mol O2/2 mol H2O)=19.46449L O2 gas
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How was island hopping a change in strategy for the us military?
Otrada [13]
Well, it was to capture Japanese controlled islands until the Japanese came into range of the American Bombers, which was different from them going after the Japanese and invading the islands.
7 0
3 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur
Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of C_5H_{12} = 21.9 g

Molar mass of C_5H_{12} = 72.15 g/mole

Molar mass of H_2O = 18 g/mole

First we have to calculate the moles of C_5H_{12}.

\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles

Now we have to calculate the moles of H_2O.

The balanced chemical reaction will be,

C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)

From the balanced reaction we conclude that

As, 1 mole of C_5H_{12} react to give 6 moles of H_2O

So, 0.3035 moles of C_5H_{12} react to give 0.3035\times 6=1.821 moles of H_2O

Now we have to calculate the mass of H_2O.

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g

Therefore, the mass of water produced will be 32.78 grams.

3 0
3 years ago
A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
aliina [53]

Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

7 0
3 years ago
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