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Scilla [17]
3 years ago
5

Please help me with chemistry. Thanks

Chemistry
2 answers:
il63 [147K]3 years ago
7 0
Constant:
Test tubes

Independent:
<span>volume of gas

Dependent:
</span>
<span>amount of H2O2 </span>
olga2289 [7]3 years ago
6 0
Hello there!

A constant stays the same so is the test tube.
An independent variable is changed to test the effects so is the volume of gas.
A dependent variable is being tested and so is the amount of H2O2.

Hope This Helps You!
Good Luck :)
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What can you infer about scanning probe microscopes?
d1i1m1o1n [39]
They are a relatively recent invention.

I hope it was helpful for you.
3 0
3 years ago
At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?
Anika [276]
The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL

Now, we use Boyle's law, that for a given amount of gas,
PV = constant

P x 6843 = 101.3 x 83776
P = 1,240 kPa
4 0
3 years ago
What mass of dinitrogen monoxide, N2O, contains the same number of molecules as 3.00 g of trichlorofluoromethane, CCl3F?
erica [24]

Answer:

0.9612 g

Explanation:

First we <u>calculate how many moles are there in 3.00 g of CCl₃F</u>, using its <em>molar mass</em>:

  • 3.00 g CCl₃F ÷ 137.37 g/mol = 0.0218 mol CCl₃F

Now, we need to calculate how many grams of N₂O would have that same number of molecules, or in other words, <em>the same amount of moles</em>.

Thus we <u>calculate how many grams would 0.0218 moles of N₂O weigh</u>, using the <em>molar mass of N₂O</em> :

  • 0.0218 mol N₂O * 44.013 g/mol = 0.9612 g N₂O
3 0
3 years ago
Part IV. Limiting Reactants! A Challenge Problem!
Alexxandr [17]

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles).  Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %

3 0
3 years ago
Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g
Andreas93 [3]
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
7 0
3 years ago
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