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2 years ago

Why is galvanized steel preferred for outdoor uses?

1 answer:

7 0

Galvanized steel is preferred for outdoor uses because it is ideal to prevent rotting/corrosion A steel will rot more quickly if it's exposed to a larger amount of oxygen and H2O , which will exist if we put it oudoor Coating the steel with additional zinc will slow down the process

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Matt is conducting an experiment where he compares the properties of water and

NikAS [45]

Answer:

sitric acid from lemon

Explanation:

there

3 0

3 years ago

Which of the following statements about the Earth’s atmosphere is true?

maksim [4K]

Carbon dioxide levels in the atmosphere by 2050 are predicted to be about double what they were before the. Industrial Revolution. ... 1% or 2% per decade, consistent with a warming atmosphere. .... Species are not yet responding to climate change because average temperatures have only increasedabout 0.8°C globally

can i get brainlest

8 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

How many miles of co2 are produced if 5.0 moles of c10h22 react with an excess of o2?

Naily [24]

The balanced equation for the above reaction is as follows;
2C₁₀H₂₂ + 31O₂ ---> 20CO₂ + 22H₂O
stoichiometry of C₁₀H₂₂ to CO₂ is 2:20
this means that for every 2 mol of C₁₀H₂₂ that reacts - 20 mol of CO₂ is formed
therefore when 5.0 mol of C₁₀H₂₂ reacts - 20/2 x 5.0 = 50 mol of CO₂ is formed
50 mol of CO₂ is produced.

3 0

2 years ago

PLZ LOOK AT ALL THE QUESTIONS. IF YOU GET THEM ALL, YOU ARE A GOD!!

Brut [27]

B cause it tells how it moves

7 0

3 years ago