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Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0.6160959 kg, 3.225 mg, and 5480.7 g. The total mass of the samples should be reported as?

Answers

Answer 1

Answer:

6.1 kg

Explanation:

To obtain the total mass of the sample, we must first express each mass of the sample in the same unit of measurement.

Since the SI unit of mass is kilogram (kg), we shall express the total mass of the samples in kilogram (kg).

This is illustrated below:

Mass of the samples are:

M1 = 0.6160959 kg

M2 = 3.225 mg

M3 = 5480.7 g.

Conversion of 3.225 mg to kg

1 mg = 1×10¯⁶ kg

Therefore,

3.225 mg = 3.225 × 1×10¯⁶

3.225 mg = 3.225×10¯⁶ kg

Conversion of 5480.7 g to kg

1000 g = 1 kg

Therefore,

5480.7 g = 5480.7 /1000

5480.7 g = 5.4807 kg

Thus, we can obtain the total mass of the samples as follow:

M1 = 0.6160959 kg

M2 = 3.225×10¯⁶ kg

M3 = 5.4807 kg

Total mass =?

Total mass = M1 + M2 + M3

Total mass = 0.6160959 + 3.225×10¯⁶ + 5.4807

Total mass = 6.096799125 ≈ 6.1 kg

Therefore, the total mass of the samples is approximately 6.1 kg.

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A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of the solution. The chemical reaction that takes place is represented by the following equation. Pb(NO3)2(s) + NaI(aq) → PbI2(s) + NaNO3(aq) How many moles of PbI2 can be produced?/Identify the limiting reactant/If the actual yield was 0.550 g of PbI2, what is the percent yield?

Answers

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield = 87.72%

Further explanation

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

moles of PbI₂
Limiting reactant
% yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

= 1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

How many s’mores can you make from the following combinations? What is the limiting reagent?

Answers

The limiting reagent and the number of S'mores produced for each of the reactions is given below:

Reaction 1. The limiting reagent is Cp; 1.6 S'mores are produced.

Reaction 2. The limiting reagent is M;2 S'mores are produced.

Reaction 3. The limiting reagent is Gc; 2.5 S'mores are produced.

Reaction 4. The limiting reagent is M;1 S'more is produced.

Stoichoimetry

Stoichiometry is the process of measuring quantitatively the mass and quantity relationships among reactants and products in a given reaction.

The equation of the reaction shows the stoichiometry between reactants and products.

For the given reaction, the equation of reaction is as follows:

1M + 2Gc + 3Cp ----> 1Gc2MCp3

where:

Gc = Graham Cracker
M = Marshmallow
Cp = Chocolate pieces
S’more = Gc2MCp3

From the equation of reaction:

1 marshmallow, 2 Graham cracker and 3 chocolate pieces are required to make 1 S'more

Calculating the number of S'mores and the limiting reactant of the given reaction

The stoichiometric equation is: 2Gc + 1M + 3Cp ----> 1Gc2MCp3

The ratio of Gc to M to Cp is 2 : 1 : 3

Reaction 1. 4 Gc + 2M + 5 Cp

The ratio of Gc to M to Cp in the reaction above is 2 : 1 : 2.5

Therefore the limiting reagent is Cp

3 Cp makes 1 S'more

5 Cp will make 5 * 1/3 S'more = 1.6 S'mores

Therefore, 1.6 S'mores are produced.

Reaction 2. 6 Gc + 2M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 3 : 1 : 4.5

Therefore the limiting reagent is M

1 M makes 1 S'more

2 Cp will make 2 * 1/1 S'more = 2 S'mores

Therefore, 2 S'mores are produced.

Reaction 3. 5 Gc + 3M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 1.6 : 1 : 3

Therefore the limiting reagent is Gc

2 Gc makes 1 S'more

5 Gc will make 5 * 1/2 S'more = 2.5 S'mores

Therefore, 2.5 S'mores are produced.

Reaction 4. 7 Gc + 1M + 6 Cp

The ratio of Gc to M to Cp in the reaction above is 7 : 1 : 6

Therefore the limiting reagent is M

1 M makes 1 S'more

Therefore, 1 S'more is produced.

The limiting reagent and the number of S'mores produced for each of the reactions is given below:

1. The limiting reagent is Cp; 1.6 S'mores are produced.

2. The limiting reagent is M;2 S'mores are produced.

3. The limiting reagent is Gc; 2.5 S'mores are produced.

4. The limiting reagent is M;1 S'more is produced.

Learn more about Stoichiometry and limiting reagents at: brainly.com/question/14222359

Fish and reptiles share which of these traits?A
They live in water,
B
They are cold blooded,
С
They do not have scales,
D
They do not have backbones,

Answers

I think it might be a or b but I’m 97% sure the answers is b

What mass of CO2 (in kilograms) does the combustion of a 16-gallon tank of gasoline release into the atmosphere? Assume the gasoline is pure octane (C8H18) and that it has a density of 0.70 g/mL.Express your answer in kilograms to two significant figures.

Answers

Mass of CO₂ = 1.3 x 10⁵ kg

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2C₈H₁₈ + 25O₂⇒ 16CO₂ + 18H₂O

16 gallon = 60566,6 ml

mass C₈H₁₈ :

$\tt mass=\rho* V\n\nmass=0.7* 60566.6=42396.62~g$

mol C₈H₁₈ :

$\tt MW=114.232~g/mol\n\nmol=(42396.62)/(114.232)=371.145$

mol CO₂ :

$\tt (16)/(2)* 371.145=2969.159$

mass CO₂ :

$\tt =2969.159* 44(MW~CO_2)=130642.996~g=1.3* 10^5~kg$

Calculate the mass % of magnesium sulfate (assume that there is a 1:1 mol ratio between sulfate and magnesium sulfate) in the original sample. Report your answer without units and use 3 sig figs, i.e. 55.23543% would be entered as 55.2

Answers

61.8 % is the mass percentage of magnesium sulphate.

Explanation:

The mass percent of individual solute or ion in a compound is calculated by the formula:

Grams of solute ÷ grams of solute + solvent × 100

mass percent of magnesium is calculated as 1 mole of magnesium having 24.305 grams/mole will have weight of 24.305 grams and 1 mole of MgSO4 will have 120.366 grams

Putting the values in the equation:

24.305 ÷ 144.671 × 100

= 16.8% of magnesium is in the mixture

The mass percentage of SO4 is calculated as

= 96.06 ÷ 216.426 × 100

= 44.38 %

The mass percentage of the mixture MgSO4 is 44.38 + 16.8 = 61.8 %

Mass percentage is a representation of the concentration of element or elements in a compound.

Zeros laced at the end of the significant number are...