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3 years ago

The infrared radiation traveling toward Earth is _____.

Chemistry

2 answers:

Lady bird [3.3K]3 years ago

7 0

The planet is the sun

Masteriza [31]3 years ago

4 0

The infrared radiation traveling toward Earth is the sun.

Good day! -Wajiha

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Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 78 mL of wa

Vaselesa [24]

Answer:

We need 1.1 grams of Mg

Explanation:

Step 1: Data given

Volume of water = 78 mL

Initial temperature = 29 °C

Final temperature = 78 °C

The standard heats of formation

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Step 2: The equation

The heat is produced by the following reaction:

Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)

Step 3: Calculate the mass of Mg needed

Using the standard heats of formation:

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)

−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg

(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required

(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg

We need 1.1 grams of Mg

7 0

3 years ago

The ccl4 formed in the first step is used as a reactant in the second step. if 2.00 mol of ch4 reacts, what is the total amount

Leni [432]

CH4 + 4 Cl2 â†’ CCl4 + 4 HCl (4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4 (4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl CCl4 + 2 HF â†’ CCl2F2 + 2 HCl (2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl 11.2 mol + 3.92 mol = 15.1 mol HCl from both steps

8 0

3 years ago

Extreme temperature changes or increased moisture speed up the weathering rate.

____ [38]

that is a true statement

6 0

3 years ago

Determine the number of 3s electrons in Na .

Scorpion4ik [409]

Answer:

Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.

3 0

1 year ago

g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution

77julia77 [94]

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles (1)

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] (2)

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

8 0

3 years ago