How many molecules are there in 0.45 moles of so3

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quester [9]

Karma sucide and hunger

5 0

3 years ago

an unknown volume of gas has a pressure of 0.50 atm and temperature of 325k. if the pressure is raised to 1.2atm and the tempera

postnew [5]

117 L. You can start by making a table to organize the information you are given. Then, you can use the formula PV/T=PV/T and plug in the numbers you have. You then solve for the missing volume. Remember that the initial pressure, temperature, and volume should be on one side of the equal sign, and the final pressure, volume, and temperature should be on the other side.

6 0

3 years ago

Phenolphthalein solution is used as an indicator of acids and bases. If added to an ammonia-based solution of window cleaner, we

leva [86]

I don’t know but look up the ph of ammonia and if it is below 7 then it is yellow and if it is above 7 then blue

3 0

3 years ago

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Assume that the test tube shown started out having 10.00 g of mercury(II) oxide. After heating the test tube briefly, you find 1

anyanavicka [17]

This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

$m_{HgO}^{consumed}=10.00g-1.35 g=8.65 g$

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

$m_{O_2}=8.65g-8.00g=0.65g$

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3 0

2 years ago

For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of car

Reil [10]

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

6 0

3 years ago

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