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mamaluj [8]
2 years ago
13

Marcus measured the masses and volumes of samples of four different substances, and he calculated their densities. The table sho

ws Marcus’s measured and calculated values. Substance Mass (g) Volume (cm3) Density (g/cm3) aluminum 5.7 2.1 2.7 copper 14.4 1.6 9.0 iron 9.5 1.2 7.9 titanium 8.6 1.8 4.8 Next, Marcus obtained another sample, which is made of one of the four substances that he had already measured. The table shows Marcus’s measured values for this unknown sample. Substance Mass (g) Volume (cm3) ? 9.5 2.1 What is the unknown sample made of?
Chemistry
1 answer:
Zina [86]2 years ago
4 0

<u>Given:</u>

Calculated density values-

Aluminum = 2.7 g/cm3

Copper = 9.0 g/cm3

Iron = 7.9 g/cm3

Titanium = 4.8 g/cm3

Unknown sample mass = 9.5 g

Sample volume = 2.1 cm3

<u>To determine:</u>

The identity of the unknown sample

<u>Explanation:</u>

'Density' is a physical parameter which can be used to identify the nature of the unknown substance.

Density = Mass/Volume

For the unknown sample

Density = 9.5 g/2.1 cm3 = 4.52 g/cm3

This matches closely with the calculated density of titanium

Ans: The unknown substance is made of titanium

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3 years ago
4. How many grams of hydrogen is produced from 12.5 g of Mg reacting with hydrochloric acid in this
Volgvan

From the reactions, 1.04 g of H2 and 7.995 g of aluminum phosphate is produced.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with the amount of substances that participates in a reaction.

For reaction 1;

Mg + 2HCl → MgCl₂ + H₂

Number of moles of Mg reacted =  12.5 g/24g/mol = 0.52 moles

If 1 mole of Mg produced 1 mole of H2

0.52 moles produces 0.52 moles of H2

Mass of H2 =  0.52 moles * 2 g/mol = 1.04 g

For reaction 2;

2Li3PO4 + Al2(SO4)3 → 3Li2SO4 + 2AIPO4

Number of moles of lithium phosphate =  7.5 g/116 g/mol = 0.065 moles

2 moles of Li3PO4 produced 2 moles of AIPO4

0.065 moles of  Li3PO4 produced 0.065 moles of AIPO4

Mass of AIPO4  = 0.065 moles  * 123 g/mol = 7.995 g

Learn more about stoichiometry:brainly.com/question/9743981

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8 0
2 years ago
What would the products of a double-replacement reaction between KBr and CaO be? (Remember: In double-replacement reactions, the
11111nata11111 [884]

ANSWER

\begin{gathered} \text{ 2KBr }+\text{ CaO }\rightarrow\text{ K}_2O\text{ }+\text{ CaBr}_2 \\ Option\text{ B} \end{gathered}

EXPLANATION

Given that;

The two reactants are KBr and CaO

Double replacement reaction is a type of chemical reaction that occur when two reactants exchange cations and anions to yield new products.

\text{ 2KBr + CaO  }\rightarrow\text{ K}_2O\text{ + CaBr}_2

Therefore, the resulting products of the given data are K2O + CaBr2

The correct answer is option B

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1 year ago
The solubility of kcl in ethanol is 0.25 g / 100 ml at 25 oc. how does this compare to the solubility of kcl in water?
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8 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
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