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3 years ago

Volume of 100ml. it’s density is 0.15g/ml. what is its mass?

Chemistry

2 answers:

11Alexandr11 [23.1K]3 years ago

3 0

$V=100mL\\
d=0,15\frac{g}{mL}\\\\
m=dV=0,15\frac{g}{mL}*100mL=15g$

Veseljchak [2.6K]3 years ago

3 0

Answer:

The mass of the substance is 15 grams.

Explanation:

Mass of the substance = m

Volume of the substance = V = 100 mL

Density of the substance = d = 0.15 g/mL

The formula of density is ;

$Density=\frac{Mass}{Volume}$

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

$0.15 g/mL=\frac{m}{100 mL}$

$m=0.15 g/mL\times 100 mL=15 g$

The mass of the substance is 15 grams.

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Answer:

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4 years ago

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3 years ago

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Answer:

The answer to your question is: letter A.

Explanation:

A Covalent bond polar is between 2 non metals where one atom is bigger than the other one so the distribution of charges creates this polarity.

A. One atom attracts shared electrons more strongly than the other atom This is the correct definition of bond polar, one element is bigger and stronger than the other element.

B. One atom has transferred its electrons completely to another atom This definition is incorrect, it is the definition of ionic bonding.

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3 years ago

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3 years ago

Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to:

$5$ $\rm O_2$ molecules.

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