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2 years ago

NaClO3 > NaCl + O2 Balance

Chemistry

1 answer:

tiny-mole [99]2 years ago

6 0

Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

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The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sa

Maksim231197 [3]

Answer:

Explanation:

a )

m = m₀ $e^{-\lambda t$

m is mass after time t . original mass is m₀ , λ is disintegration constant

λ = .693 / half life

= .693 / 1590

= .0004358

m = m₀ $e^{- 0.0004358 t}$

b )

m = 50 x $e^{-.0004358\times 500}$

= 40.21 mg .

c )

40 = 50 $e^{-.0004358t$

.8 = $e^{-.0004358t$

$e^{.0004358t$ = 1.25

.0004358 t = .22314

t = 512 years .

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2 years ago

(b) What amount (mol) of H₂ can be produced from the given mass of H₂O?

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8 moles of H 2O are produced.

First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.

<h3>How many moles of H2 are required to make one mole of NH3?</h3>

Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.

<h3>How many moles of h2 are needed to produce a solution?</h3>

An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).

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1 year ago

Sodium (na), atomic number 11, has a tendency to lose an electron in the presence of chlorine. after losing the electron, na has

Paha777 [63]

Eleven because protons don’t leave.

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3 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

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3 years ago

Draw the structure of the major organic product isolated from the reaction of 3-hexyne with hydrogen (1 mol), Lindlar palladium.

ohaa [14]

Answer:

See explanation and image attached

Explanation:

When 1 mole of hydrogen is added to 3-hexyne, the addition occurs on the same face or side of the triple bond to yield cis-2-hexene in the presence of lindlar catalyst which prevents further hydrogenation. This is known as syn addition.

The syn addition is achieved because the hydrogen molecule is first adsorbed on the surface of the palladium metal and is subsequently attached to the same face of the triple bond in 3-hexyne as shown in the image attached to this answer.

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2 years ago