Identify the reactant and products in the following chemical reaction: CH4 + 2O2 ➡ CO2 + 2H2O

Chemistry

1 answer:

kap26 [50]3 years ago

6 0

The reactants are methane and oxygen.

The products are carbon dioxide and water.

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How many particles are in 12.47 grams of NaCl?

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If we are talking about moles then the answer to that is 0.22

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During a lab, you heat 1.62 g of a CoCl2 hydrate over a Bunsen burner. After heating, the final mass of the dehydrated compound

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Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
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A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 8.5 oz . aluminum has a density of 2.70 g/cm3. w

Serhud [2]

Hello!

We have the following data:

Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)

To move on, we must transform the area of 50 ft² in cm², let's see:

1 ft² ------- 929,0304 cm²
50 ft² ----- A

$A = 50*929,0304$

$\boxed{A = 46451,52\:cm^2}\Longleftarrow(Area)$

In the same way, we will convert the mass of 8.5 oz in grams, see:

1 oz -------- 28,3495 g
8,5 oz ------- m

$m = 8,5*28,3495$

$\boxed{m = 240,97075\:g}\Longleftarrow(mass)$

Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{240,97075\:\diagup\!\!\!\!\!g}{2,70\:\diagup\!\!\!\!\!g/cm^3}$

$V = 89,24842593... \to \boxed{V \approx 89,25\:cm^3}$

The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)

In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:

$V (volume) = A (Area) * T (Thickness)$

$89,25\:cm^3 = 46451,52\:cm^2\:*\:T$

$46451,52\:cm^2*T = 89,25\:cm^3$

$T = \dfrac{89,25\:cm^3}{46451,52\:cm^2}$

$T = 0,001921358009...\:cm$

We will convert to millimeters, going through a decimal place on the right

$T = 0,01921358009..\:mm$

$\boxed{\boxed{T \approx 0,0192\:mm\:or\:T\approx \:1,92*10^{-2}\:mm}}\end{array}}\Longleftarrow(thickness)\qquad\checkmark$

Hope this helps! :))

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3 years ago