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2 years ago

How would a sandblasted rock differ from a rock that hasnt been sandblasted

1 answer:

7 0

Sandblasted: More smooth because all the rough edges are "blasted" away. And slightly smaller because of the loss of the edges.

Non-Sandblasted: Not smooth; rough.

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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

2 years ago

Reactivity and flammability are are examples of properties.

nika2105 [10]

Answer:

they are eamples of chemical properties

Explanation:

flammibility is the chemical makeup of the element that makes it flamable and reactivitie is the chemical makeup of an element that is able to reacte to another element

3 0

2 years ago

HNO3 + H2O → H3O^+ + NO3 which ones are acids and which ones are bases

nlexa [21]

Answer:

HNO₃ and H₃O⁺ are acids

H₂O and NO₃⁻ are bases

Explanation:

The chemical equation is:

HNO₃ + H₂O → H₃O⁺ + NO₃⁻

There are several definitions of acid and bases: Arrhenius', Bronsted-Lowry's and Lewis'.

Bronsted-Lowry model defines and acid as a donor of protons, H⁺.

In the given equation HNO₃ is such substance: it releases an donates its hdyrogen to form the H₃O⁺ ion.

On the other hand, a base is a substance that accepts protons.

In the reaction shown, H₂O accepts the proton from HNO₃ to form H₃O⁺.

Thus, H₂O is a base.

In turn, on the reactant sides the substances can be classified as acids or bases.

H₃O⁺ contain an hydrogen that can be donated and form H₂O; thus, it is an acid (the conjugated acid), and NO₃⁻ can accept a proton to form HNO₃; thus it is a base (the conjugated base).

4 0

2 years ago

How are hail and snow similar?

Debora [2.8K]

Answer:

the answer is a

Explanation:

i did process of elimination and i got a

3 0

2 years ago

Hydrogen sulfide is composed of two elements: hydrogen and sulfur. in an experiment, 6.500 g of hydrogen sulfide is fully decomp

LiRa [457]

Hydrogen sulfide = hidrogen + sulfur

6.500 g

a) 0.384 g + x

=> 6.500 = 0.384 + x => x = 6.500 - 0.384 = 6.116 g

Answer: 6.116 g of sulfur must be obtained

b) this experiment demonstrate the conservation of mass.

c) Dalton's atomic model states that the atoms cannot be created, split or be destroyed, and so in a chemical reaction the atoms rearrange but the number of each type of atoms remain constant, so the mass of each type of atoms and the total mass remain constant.

8 0

3 years ago