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3 years ago

"moves as a current" is it conduction radiation or convection"

Chemistry

1 answer:

Cloud [144]3 years ago

8 0

Conduction tranfer with the touche
Radiation transfer in distance
And convection moves as a current

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Ice melts to form water. And water freezes to form ice. Choose the TRUE statement about these changes.

Dmitrij [34]

Answer:

A. Both freezing and melting are physical changes.

Explanation:

Even if you were to freeze water, the molecules are still water molecules, and vise versa with melting it.

4 0

3 years ago

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Which of the outer planets is the biggest A.jupiter B.saturn C.your anus D.neptune

Alina [70]

A. Jupiter. is the correct answer. Mark as brainliest please.

5 0

2 years ago

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Give the number of significant figures: 369,132,000 sec

Yuliya22 [10]

The number of significant figures in 369,132,000 is 6

3 0

2 years ago

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As part of your job you are asked to make 1 liter of a 0.5 molar sucrose solution. how much sucrose (c12h22o11) do you need? use

kakasveta [241]

Answer:- 171 g

Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.

We know that molarity is moles of solute per liter of solution.

If molarity and volume is given then, moles of solute is molarity times volume in liters.

moles of solute = molarity* liters of solution

moles of solute = 0.5*1 = 0.5 moles

To convert the moles to grams we multiply the moles by molar mass.

Molar mass of sucrose = 12(12) + 22(1) + 11(16)

= 144 + 22 + 176

= 342 grams per mol

grams of sucrose required = moles * molar mass

grams of sucrose required = 0.5*342 = 171 g

So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.

6 0

3 years ago