All categories

2 years ago

How many quarters fit in a 1.75 liter bottle?

Chemistry

2 answers:

WARRIOR [948]2 years ago

8 0

1.85 quarts can fit into a 1.75 liter bottle

Volgvan2 years ago

3 0

Answer:

In a 1.75 liter bottle fit 1.8492 quarters.

Explanation:

The quarter is a unit of volume. It is equivalent to $\frac{1}{4}$ of a gallon :

$1gallon=3.78541L$

Where L is liter

⇒

$\frac{1}{4}gallon=1quarter=\frac{3.78541L}{4}$

To find how many quarters fit in a 1.75 L bottle we can write :

$\frac{\frac{3.78541L}{4}}{1quarter}=\frac{1.75L}{x}$

Solving for x :

$x=\frac{(1.75L).(1quarter)}{\frac{3.78541L}{4}}$

$x=1.8492quarters$

You might be interested in

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: