Which of the following does nitrogen have an oxidation state of +4?

Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.

BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Reading the Periodic Table
Writing Compounds
Using Dimensional Analysis

Explanation:

Step 1: Define

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

Step 2: Define conversions

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

Step 3: Use Dimensional Analysis

$6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} )$ = 511.901 g NO₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

511.901 g NO₂ ≈ 510 g NO₂

6 0

3 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

C2952H4664N812O832S8Fe4+NaC4H3O4SAu+Fe(SCN)2+Fe(NH4)2(SO4)2.6H2O+C4H8Cl2S+C8H12MgN2O8=C55H72MgN4+Na4Fe(CN)6+AuSC6H11O5+HClO4+H2S

Scorpion4ik [409]

oh dannnnnnnnnnnnnnnnnnnnnnnnnng XD

7 0

3 years ago

Which member of each pair is more soluble in diethyl ether? Why? (c) MgBr₂(s) or CH₃CH₂MgBr(s)

White raven [17]

CH3CH2MgBr is more soluble in diethyl ether .

We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .

Hence the compound with similar interaction can dissolve in diethyl ether .

Here , MgBr2 is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .

but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .

thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .

Learn more about polar solvent here :

brainly.com/question/3184550

#SPJ4

4 0

2 years ago

What five things is the element carbon a part of?

Ivanshal [37]

Answer:

oceans,air,rocks,soil, also living things

Explanation:

4 0

3 years ago