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3 years ago

How many grams of sucrose are in 1.80 of a 0.718 sucrose solution

Chemistry

1 answer:

Ronch [10]3 years ago

8 0

Answer:

445 g

Explanation:

Data Given:

Volume of sucrose solution = 1.8 L

Molarity of sucrose solution = 0.718 M

mass of solute = ?

Solution:

First we will find mole of solute in a Molar Solution of sucrose

Formula used

Molarity = No. of moles of solute / of solution in L

where

solute is sucrose

Put Values in above equation

0.718 M = No. of moles of sucrose / 1.8 L

Rearrange the equation as we have to find moles of sucrose

No. of moles of sucrose = 0.718 x 1.8

No. of moles of sucrose = 1.3 mole

Now,

we have to find mass of sucrose

For this formula will use

mass in grams = no. of moles x Molar mass . . . . . .(1)

Formula of Sucrose = C₁₂H₂₂O₁₁

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

put values in equation 1

mass in grams = 1.3 mol x 342 g/mol

mass in grams = 444.6 g

Round the figure. So,

mass of C₁₂H₂₂O₁₁ (Sucrose) = 445 g

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OLga [1]

Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.

Another way to look at this problem is to use Newton's second law of motion. The first law states that $F = m\times a$ , where $a$ is the acceleration $F$ is the net force and $m$ is the mass of the object.

When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.

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3 years ago

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

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3 years ago

How many moles are in 17.50g of SO2

monitta

The formula for mole is
n= Mass/Mol mass
Mol Mass: S=32
O2= 16(2)
—————
64 g/mol
N= 17.50 g
————— (cancel both g)
64 g/mol
= 0.27 mol is the answer

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3 years ago

Part 1: Name the type of chemical reaction that occurs when calcium hydroxide (Ca(OH)2) reacts with nitric acid (HNO3).

liberstina [14]

Answer:

For 1: Neutralization reaction

For 2: Zinc is more reactive than lead and less reactive than calcium.

Explanation:

For (1):

When a base reacts with an acid to form a salt and water molecule, it is known as a neutralization reaction. The general equation follows:

$HX+BOH\rightarrow BX+H_2O$

The chemical equation for the reaction of calcium hydroxide and nitric acid follows:

$Ca(OH)_2(aq)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+2H_2O(l)$

For (2):

A single displacement reaction is defined as the reaction in which a more reactive metal displaces a less reactive metal from its salt solution. The general chemical equation follows:

$A+BX\rightarrow AX+B$