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schepotkina [342]

2 years ago

15

A bottle of antiseptic hydrogen peroxide (H2O2) is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solutio

n?
Use this formula: Percent by volume: volume of solute/volume of solution×100%

1 answer:

PolarNik [594]2 years ago

6 0

Answer:

= 12 mL H202

Explanation:

Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v

It implies that, 3ml H2O2 is present in 100 ml of solution.

Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;

we have;

(3.0 mL/ 100 mL) × 400 mL

= 12 mL H202

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Determine the y component of reaction at a using scalar notation. express your answer to three significant figures and include t

blagie [28]

In order to compute the y-component of a vector, we simply use the formula:

Fy = F*sin(∅)
Where ∅ is the angle of the vector measured from the positive x-axis and F is the magnitude of the vector.

Similarly, the x-component is calculated by substituting sin(∅) with cos(∅)

6 0

2 years ago

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 174 g

$T_{final}$ = final temperature of water = 317.4 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0

2 years ago

Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is

Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

no. of moles of NaCl = mass of NaCl / molar mass

no. of moles of NaCl = 0.45g / 58.44 g/mol

no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

M = moles of solute / Liter of solution . . . . . . . . . . (1)

M = 0.0077 g / 0.1 L

M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0

2 years ago

How many total ions are present in 347g of cacl2?

vladimir1956 [14]

In 1 mole of $CaCl_{2}$ , there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of $Cl^{-1}$ .

$CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}$

Molar mass of $CaCl_{2}$ is 110.98 g/mol. Calculating number of moles from given mass as follows:

$n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol$

Thus, number of moles of ions will be $3\times 3.12 mol=9.38 mol$ .

Since, 1 mole of any substance has $6.023\times 10^{23}$ units of that substance where $6.023\times 10^{23}$ is Avogadro's number.

Thus, 9.38 mol of ions will have $9.38\times 6.023\times 10^{23}=5.65\times 10^{24}$ number of ions.

Therefore, total number of ions in 347 g of $CaCl_{2}$ is $5.65\times 10^{24}$ .

8 0

2 years ago

A LIQUID HAS THE FOLLOWING PROPERTIES: ONE PHASE, COLORLESS, BOILING POINT 95 TO 97 oC. WHICH OF THE FOLLOWING BEST DESCRIBES A

ololo11 [35]

I think it is D. coarse mixture
hope this helps:)

8 0

2 years ago