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3 years ago

Which is a diatomic molecule? Na-CI o=O H-F Si-O which one

Chemistry

2 answers:

pshichka [43]3 years ago

8 0

The diatomic molecule is H-F!

allochka39001 [22]3 years ago

7 0

Is “o=O” suppose to be “O-O” because if so then that would be the answer as it would represent O2 ( oxygen )

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QUESTION 21 The combustion of ammonia in the presence of excess oxygen yields NO 2 and H 2O: 4 NH 3 (g) 7 O 2 (g) 4 NO 2 (g) 6 H

Harrizon [31]

Answer:

The answer to your question is 47.44 g of Oxygen

Explanation:

Data

mass of Ammonia = 14.4 g

mass of Oxygen = ?

Balanced chemical reaction

4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O

Process

1.- Calculate the molar mass of Ammonia

NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g

2.- Calculate the molar mass of Oxygen

O₂ = 7[16 x 2] = 7[32] = 224 g

3.- Use proportions to calculate the mass of Oxygen

68g of NH₃ --------------------- 224 g of O₂

14.4 g of NH₃ ----------------- x

x = (14.4 x 224) / 68

x = 3225.6/ 68

x = 47.44 g

5 0

3 years ago

In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

elena55 [62]

Answer:

In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.

Explanation:

The balanced reaction is

H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.

The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:

$Molarity=\frac{number of moles}{volume}$

in units of $\frac{moles}{liter}$

then the number of moles can be calculated as:

number of moles= molarity* volume

You have acid H₂SO₄

35.00 mL= 0.035 L (being 1,000 mL= 1 L)
Molarity= 0.737 M

Then:

number of moles= 0.737 M* 0.035 L

number of moles= 0.0258

So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?

$moles of KOH=\frac{0.0258moles of H_{2} SO_{4}*2 moles of KOH }{1mole of H_{2} SO_{4}}$

moles of KOH= 0.0516

Then 0.0516 moles of KOH are needed. So you know:

Molarity= 0.827 M
number of moles= 0.0516
volume=?

Replacing in the definition of molarity:

$0.827 M=\frac{0.0516 moles}{volume}$

Solving:

$volume=\frac{0.0516 moles}{0.827 M}$

volume=0.0624 L= 62.4 mL

<u><em>In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.</em></u>

4 0

3 years ago

Here is the last two clues out of the crossword

balandron [24]

FIRST ONE IS SPRING ****
Second one is Buoyancy.

7 0

3 years ago

How many grams are in 0.220 mol of Ne?

yawa3891 [41]

Answer:

Mass = 4.44 g

Explanation:

Given data:

Number of moles of Ne = 0.220 mol

Mass in gram = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of Ne = 20.2 g/mol

by putting values,

0.220 mol = mass/ 20.2 g/mol

Mass = 0.220 mol × 20.2 g/mol

Mass = 4.44 g

5 0

3 years ago

One gram-mole of methyl chloride vapor is contained in a vessel at 100°C and 10atm.(a) Use the ideal-gas equation of state to es

777dan777 [17]

Answer:

a) The volume of the system is 3.062 Liters.

b) Percentage error results from assuming ideal-gas behavior is 9.36%.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = $10 atm$

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 100°C = 100+273K=373 K

Putting values in above equation, we get:

$(10 atm)\times V=1 mol\times (0.0821L.atm/mol.K)\times 373 K$

$V=\frac{1 mol\times (0.0821L.atm/mol.K)\times 373 K}{10 atm}$

V = 3.062 L

The volume of the system is 3.062 Liters.

Volume of the container = V' = 2.8 L

System volume = V = 3.062 L

Percentage error of volume:

To calculate the percentage error, we use the equation:

$\%\text{ error}=\frac{|\text{Experimental value - Accepted value}|}{\text{Accepted value}}\times 100$

$=\frac{|3.062L-2.8 L|}{2.8 L}\times 100=9.36\%$

Percentage error results from assuming ideal-gas behavior is 9.36%.

6 0

3 years ago