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2 years ago

D

Chemistry

1 answer:

boyakko [2]2 years ago

4 0

Answer:

My conclusion about the picture above. Is cleaning or arrange all things in the kitchen is one of the reason to avoid us in danger.The things that we can prevent danger in the kitchen always clean the kitchen room,things especially to a hazard things.

burning is one of danger things that happrn in the kitchen.

You might be interested in

Why is water said to be universal solvent?

Pachacha [2.7K]

Answer:

Explanation:

And, water is called the "universal solvent" because it dissolves more substances than any other liquid. This allows the water molecule to become attracted to many other different types of molecules.

8 0

2 years ago

Read 2 more answers

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with

emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

It is a stichiometry problem.

The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,

Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,

where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,

V is the volume of the gas in L <em>(V = 7.80 L)</em>,

n is the number of moles in mole,

R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,

T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

<em>Using cross multiplication:</em>

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0

3 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.

patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of $NH_{3}$ is as follows:.

$2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)$

And, initially only $NH_{3}$ is present.

The given data is as follows.

$P_{NH_{3}}$ = 2.3 atm at equilibrium

$P_{H_{2}} = 3 \times P_{N_{2}}$ = 0.69 atm

Therefore,

$P_{N_{2}} = \frac{0.69 atm}{3}$

= 0.23 aatm

So, $P_{NH_{3}}$ = 2.3 - 2(0.23)

= 1.84 atm

Now, expression for $K_{p}$ will be as follows.

$K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}$

$K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}$

= $\frac{0.23 \times 0.33}{3.39}$

= 0.0224

or, $K_{p} = 2.2 \times 10^{-2}$

Thus, we can conclude that the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is $2.2 \times 10^{-2}$ .

6 0

3 years ago

Sulfur is a nonmetalic mineral that has a density of 2 g/cm3. The volume of a sample of sulfur was measured to be 5.0 cm3. what

taurus [48]

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 2 × 5

We have the final answer as

Hope this helps you

8 0

2 years ago

Calculate the theoretical carbonaceous and nitrogenous oxygen demand for:

serg [7]

Answer:

The correct answer is 129 mg and 232 mg.

Explanation:

Theoretical carbonaceous oxygen demand:

The reaction will be,

C₂H₆O₂ + 5/2 O₂ ⇒ 2CO₂ + 3H₂O

Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.

The molecular mass of ethylene glycol is 62 grams per mole.

The given mass of ethylene glycol is 100 mg or 0.1 grams

The moles of ethylene glycol will be,

Moles = Weight/Molecular mass

= 0.1/62 = 1.613 × 10⁻³ mol

For 1.613 × 10⁻³ mol, the moles of O₂ will be,

= 2.5×1.613×10⁻³

= 4.0.×10⁻³ × 32mol

= 0.129 grams or 129 mg.

The theoretical nitrogenous oxygen demand is:

The reaction will be,

2NH₃-N + 9/2O₂ ⇒ 4HNO2 + H₂O

Thus, for 2 moles of NH₃-N, 4.5 moles of O₂ is needed,

Therefore, for 1 mol of NH₃-N, the oxygen required will be,

= 4.5/2 = 2.25 mol

The given mass of NH₃-N is 100 mg, the moles of NH₃-N will be,

Moles = 100×10⁻³/31 = 3.225 × 10⁻³ mol (The molecular mass of NH₃-N is 31 gram per mole)

The moles of O₂ is 2.25 × 3.225 × 10⁻³ = 7.258 × 10⁻³ mol.

Now the mass of O2 will be,

= 7.258 × 10⁻³ × 32

= 0.232 grams

= 232 mg

5 0

2 years ago