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Luda [366]
3 years ago
15

Make a list of silver current industrial uses. Try to include the approximate annual demand if possible. Create a table to displ

ay this information.
Chemistry
2 answers:
r-ruslan [8.4K]3 years ago
3 0

Answer:

Los usos de la plata son cientos, sobre todo en procesos industriales, comerciales y hasta personales. Su resistencia a la corrosión la hace ideal para la elaboración de recipientes especiales o para recubrir otros metales.

Explanation:

dezoksy [38]3 years ago
3 0

Answer:

Here's the latest information I can find  

Explanation:

\begin{array}{lr}&\textbf{Annual demand}\\\textbf{Use} & \textbf{/million ounces} \\\text{Electrical/electronics} & 248.5 \\\text{Solar panels} & 98.7 \\\text{Brazing/soldering} & 58.0 \\\text{Photography} & 39.3 \\\text{Ethylene oxide} & 5.4 \\\text{Other$^{\ast}$} & 146.9 \\\end{array}

* Other industrial uses include pharmaceutical and medical, industrial catalysts, mirrors and windows, nanotechnology, superconductors, and many others.

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The chemical equation for the formation of ammonia is unbalanced.
ratelena [41]
Balance Chemical equation is as follow,

<span>                           3 H</span>₂ <span>(g) + N</span>₂ <span>(g)    </span>→<span>    2 NH</span>₃ <span>(g)

According to balanced equation, 3 Molecules (3 moles) of Hydrogen reacts with 1 Molecule of N</span>₂ to produce 2 moles (2 Molecules) of NH₃.

Result:
          2 Molecules of Ammonia are produced by reacting 3 molecules of Hydrogen and 1 molecule of Nitrogen.
7 0
3 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

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
Volume of water displaced = 0.175 L Mass g Density g/cm3 Hardness: = Determine the unknown mineral. Unknown mineral: 7​
erastovalidia [21]

Answer:

Mass: 981.0 g

Density: 5.61 g/cm^3

Hardness: = 2.5 - 3

Unknown material: Chalcocite

Explanation:

5 0
3 years ago
Americans combined drive about 4.0 x 109 kilometers a day and get an average of 20 miles per gallon of gasoline. For each kilogr
Nimfa-mama [501]

Answer:

303,882.84649 kg\times 3=9.12\times 10^5 kg of carbon dioxide gas.

Explanation:

Average distance covered by Americans in a day= 4.0\times 10^9 km

1 day = 24 × 60 min = 1,440 min

Average distance covered by Americans in a minute= \frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km

Average mileage of the car = 20 miles/gal = 32.18 km/gal

1 mile = 1.609 km

20 miles = 20 × 1.609 km = 32.18 km

Volume of gasoline used in minute = \frac{2,777,777.78 km}{32.18 km/gal}

V=86,320.00 gal

V=86,320.00\times 3.7854 L

(1 L = 1000 mL)

V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL

Mass of 86,320.00 gallons of gasoline = m

Density of the gasoline = d = 0.93 g/cm^3=0.93 g/mL

1 mL= 1 cm^3

m=d\times V=0.93 g/mL\times 326,755,748.91 mL

m=303,882,846.49 g=303,882.84649 kg

1 kilogram of gasoline gives 3 kg of carbon dioxde gas .

Then 303,882.84649 kg of gasoline will give :

303,882.84649 kg\times 3=9.12\times 10^5 kg of carbon dioxide gas.

8 0
3 years ago
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