When the equation H202 + H2O + O2 is completely balanced, the SUM of all the coefficients will be

what is the main 'impurity' collected with the oil in distillation. PLEASE HELP IM IN END OF YEAR EXAM. II WILL GIVE BRAINLIEST

Zigmanuir [339]

Answer:

I think it is sulfur

Explanation:

5 0

3 years ago

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati

Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt + Ni(NO₃)₂ = No reaction.

8 0

3 years ago

15.0 mL of an unknown clear liquid is added to a 50 mL graduated cylinder. The mass of the liquid is determined to be 12.7 grams

sveticcg [70]

Answer:

$\boxed {\tt A. \ d=0.85 \ g/mL}$

Explanation:

Density is found by dividing the mass by the volume.

$d=\frac{m}{v}$

The mass of the liquid is 12.7 grams.

We know that 15 mL of this liquid was added to a 50 mL graduated cylinder. Therefore, the volume is 15 mL. The 50 mL is not relevant, it only tells us about the graduated cylinder.

$m= 12.7 \ g\\v= 15 \ mL$

Substitute the values into the formula.

$d=\frac{12.7 \ g}{ 15 mL}$

Divide.

$d=0.846666667 \ g/mL$

Round to the nearest hundredth. The 6 in the tenth place tells us to round the 4 to a 5.

$d \approx 0.85 \ g/mL$

The density of the liquid is about 0.85 grams per milliliter and choice A is correct.

4 0

3 years ago

When using ion-selective electrodes, to compensate for a complex or unknown matrix, the _____________ method can be used to dete

VMariaS [17]

When using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration. Option D

<h3>What are ion-selective electrodes?</h3>

Analytical chemistry is a science that deal with the measurement and detection of the accurate amount of a substance. Analytical chemistry plays a large role in environmental management as it helps in the determination of the levels of contaminants in a sample.

An ion selective electrode is used in analytical chemistry to measure the amount of a target ion by converting its activity into a measurable electrical signal.

Hence, when using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration.

Learn more about ion-selective electrodes:brainly.com/question/14987024

#SPJ1

7 0

2 years ago

S8 + 24 F2 ⟶ 8 SF6

Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆ to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.

____________________

Here's another example just for grins ...

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

Given: 253g 307g ? ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced

Limiting Reactant

moles C₂H₆O = 253g/46g/mol = 5.5 moles => 5.5/1 = 5.5

moles O₂ = 307g/32g/mol = 9.6 moles =><em> 9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

------------ 9.6 mole (L.R.) ? ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0

3 years ago