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1 year ago

How many liters of cane juice is needed to supply 5g sucrose if cane juice contains 12% sucrose

Chemistry

1 answer:

pickupchik [31]1 year ago

6 0

Given parameters:

Mass of sucrose = 5g

Density of sucrose = 1.12g/mL

Percentage of sucrose per liter of cane juice = 12%

Unknown:

Volume of cane juice needed = ?

We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.

Mathematically;

Density = $\frac{mass}{volume}$

Now solve for the volume of sucrose;

1.12g/mL = $\frac{5}{Volume}$

Volume = $\frac{5}{1.12}$ = 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L

Since 12% of 1 liter of cane juice is sucrose;

12% of x liter of cane juice = 4.46 x 10⁻³L

Volume of cane juice = 4.46 x 10⁻³ x $\frac{100}{12}$ = 0.037L

Volume of cane juice is 0.037L

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solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

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No of portions n=3

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2 years ago

Find the mass in grams of 1.40x10^23 molecules of n2

mina [271]

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1 year ago

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lara31 [8.8K]

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2 years ago

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Assume the atomic mass of element X is 22.99 amu. A 17.15−g sample of X combines with 14.17 g of another element Y to form a com

Alenkasestr [34]

The formula of the compound is XY. This means that the relation between the moles is 1: 1. One mole of X per one mole of Y.

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# of moles of X = weigth of X / atomic mass of X = 17.15 g / 22.9 g/mol = 0.74598

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1 year ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

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