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Ivenika [448]
3 years ago
11

Hydrogen peroxide is sold as a 3.0% (by mass) solution. The rest of the solution is water . How many grams of hydrogen peroxide

are in 250 grams of this solution?
Chemistry
1 answer:
jolli1 [7]3 years ago
4 0
___3% H2O2 (mass/vol) 

<span>a) mass bought = 3g/100ml * 250 ml = ?? g H2O2 </span>

<span>b) moles / 0.250 liter = (?? g H2O2 / MW H2O2 g/mole) / 0.250 liter = ??M </span>

<span>Plug and SOLVE </span>

<span>Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.

</span>
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Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh
olga55 [171]
<span>1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
3 0
3 years ago
What happens to temperature when the frequency is increased and when the frequency is decreased
Ymorist [56]

Answer:

As the frequency increases, the wavelength decreases. ... (a) For a given sound, as the temperature increases, what happens to the frequency? There is no change in frequency. The speed of sound increases by about 0.5 m/s for each degree Celsius when the air temperature rises.

5 0
3 years ago
Assume that the NO, concentration in a house with a gas stove is 150 pg/m°. Calculate the equivalent concentration in ppm at STP
jok3333 [9.3K]

Explanation:

It is known that for NO_{2}, ppm present in 1 mg/m^{3} are as follows.

                      1 \frac{mg}{m^{3}} = 0.494 ppm

So, 150 pg/m^{3} = \frac{150}{1000} mg/m^{3}

                       = 0.15 mg/m^{3}

Therefore, calculate the equivalent concentration in ppm as follows.

             0.15 \times 0.494 ppm

              = 0.074 ppm

Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.  

4 0
3 years ago
How many moles of gas would occupy a 25.0 liter when the temp is 22.0 degrees celsius and the pressure is 646 torr
Luda [366]

Answer:

0.877 mol  

Step-by-step explanation:

We can use the<em> Ideal Gas Law </em>to solve this problem.

pV = nRT     Divide both sides by RT

 n = (pV)/(RT)

Data:

p = 646 torr

V = 25.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 22.0 °C

Calculations:

(a) <em>Convert the pressure to atmospheres </em>

p = 646 torr × (1 atm/760 torr) = 0.8500 atm

(b) <em>Convert the temperature to kelvins </em>

T = (22.0 + 273.15) K = 295.15 K

(c) <em>Calculate the number of moles </em>

n = (0.8500 × 25.0)/(0.082 06 × 295.15)

  = 0.877 mol

3 0
3 years ago
Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0
3 years ago
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