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Naddika [18.5K]
2 years ago
8

O1S. Answer the following questions using Boyle's law:

Chemistry
1 answer:
Olenka [21]2 years ago
3 0

Answer: 0.5 atm

Explanation:

you would want to use Boyles law which is P1P2=P2V2

or in this case, 1.00atm x 5.00L=P2 x 10.0L

                       1.00x5.00/10.00= 0.50atm

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The usual units of density are : <br> cm3/ g<br> cm2/g<br> g/cm<br> Nm
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The usual units of density are g/cm.

You may have also seen g/mL used for density. Keep in mind that 1 cm = 1 mL.
3 0
3 years ago
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Percent yield refers to the relationship between the predicted amount of product
dsp73
True because it actually formed I know
5 0
2 years ago
Natation TT mean to genetics
neonofarm [45]
I am not sure what you are asking, but TT likely refers to the pure tall peas Mendel worked with.
4 0
2 years ago
How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration
Elena L [17]

Answer:

Amount of HCL = 0.00318 L  of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH  = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH  = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed =  0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795  / 2.5

Amount of HCL = 0.00318 L  of 3.18 ml

4 0
3 years ago
Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)
vampirchik [111]

Answer: Oxygen is the limiting reagent.

Explanation: CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)

As can be seen from the given balanced equation:

3 moles of O_2 reacts with 1 mole of CS_2

1.52 moles of O_2 reacts with=\frac{1}{3}\times 1.52=0.51moles of CS_2

Thus O_2 is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of CS_2 will remain as such and thus  CS_2 is an excess reagent.

6 0
3 years ago
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