What is the use of oxyacetylene flame

Chemistry

2 answers:

harkovskaia [24]3 years ago

7 0

Answer:

An oxyacetylene torch can also be used for welding. When welding with an oxyacetylene torch, the flame is used to produce molten metal along the edge of two work pieces.

hope this helped

if you want plz give brainliest to help you more

statuscvo [17]3 years ago

4 0

Simple answer is

The oxyacetylene flame is a gas torch that burns a mixture of acetylene and oxygen to produce a high-temperature flame (3,000°C or 5,400°F) that can be used for welding, cutting steel, and other metal working.

You might be interested in

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$