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4 years ago

What happens to an electromagnetic wave as it passes from space to matter

1 answer:

8 0

When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.

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Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2

Helen [10]

Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = 23350L de O₂

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) = 3.33 tanques de O₂

6 0

3 years ago

39. Analyze What subscripts would you most likely use if

Igoryamba

Based on their valencies, the subscripts of the ionic compounds formed will be:

1 and 1
2 and 1
1 and 2
2 and 2

<h3>What subscripts would you most likely use if the following substances formed an ionic compound?</h3>

A. An alkali metal and a halogen.

An alkali metal and a halogen both have valencies of one.

Therefore the subscripts would be 1.

B. An alkali metal and a non-metal from group 16.

An alkali metal has a valency of 1 and a group 16 non-metal has a valency of 2.

By exchange of valencies, the subscript would be 2 and 1.

C. An alkaline earth metal and a halogen.

An alkaline earth metal has a valency of 2 and a halogen has a valency of 1.

By exchange of valencies, the subscripts would be 1 and 2.

D. An alkaline earth metal and a non-metal from group 16.

An alkaline earth metal has a valency of 2 and a non-metal from group 16 has a valency of 2.

By exchange of valencies, the subscripts would be 2 and 2.

Therefore, the subscripts of the ionic compounds formed will be:

1 and 1
2 and 1
1 and 2
2 and 2

Learn more about about valency at: brainly.com/question/2284519

4 0

3 years ago

Can someone please help me and don’t answer if you don’t know the answer please thank you :)

o-na [289]

Combustion means the burning of something. So in the given question, you could say, “In this question, combustion means the burning of methane”.

8 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago

A student increases the temperature of a 200 cm3 balloon from 60⁰C to 180⁰C. What should the new volume of the balloon be? Use t

lara [203]

236 cm3 Is the correct answer..

5 0

3 years ago

Read 2 more answers