Answer:
2 hours Speed =110/2=55 miles ler hour
Explanation:
atomic hydrogen torch utilizes an electric arc whereby two closely - but not touching electrodes - result in the release of powerful electric spark as the current tries to flow through the gap. The gap is filled with hydrogen gas in an atomic hydrogen torch rather than air. The electric arch is split the hydrogen gas molecules into hydrogen atoms (some in plasma form). When the hydrogen atoms land on cooler objects like the metal being welded or cut, they region back to H₂ molecules releasing enormous amounts of heat on the surface. Surface temperatures can reach 4000 °C. The use of hydrogen gas protects the metal being welded from oxidation. Oxidation may compromise the quality of the weld.
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The concentration of the original calcium ions is 0.005 M
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.
As such we have the equation;
Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)
Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol
= 0.0022 moles
Now;
1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.
Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L
= 0.005 M
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Answer:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
Explanation:
The standard formation reaction is the synthesis of 1 mole of a substance from its elements in their most stables forms under standard conditions. The balanced chemical equation is:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
