All categories

2 years ago

Give the reason why the empty crucible should be heated before starting the experiment

1 answer:

3 0

To make sure no cracks are in the crucible and also to remove any moisture present in the crucible by the process of heating.

You might be interested in

Which sequence represents the relationship between temperature and volume as explained by the kinetic-molecular theory? higher t

RSB [31]

Answer:

Explanation:

2020edge

7 0

3 years ago

Read 2 more answers

Pls help fast!!! Test due soon!

Firlakuza [10]

Answer:

Explanation:

your wellcome

4 0

3 years ago

Read 2 more answers

Predict the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S

777dan777 [17]

1) Chemical equation

16Fe(s) + 3S8(s) ---> 8Fe2S3

2) Molar ratios:

16 mol Fe : 3 mole S8 : 8 mol Fe2S3

3) Convert masses in grams to number of moles

number of moles = mass in grams / molar mass

a) iron, Fe

mass = 3.0 g
atomic mass = 55.845 g/mol

=> number of moles of Fe = 3.0g / 55.845 g/mol = 0.0537 mol

b) Sulfur, S8

mass = 2.5 g
molar mass = 8*32.065 g/mol = 256.52 g/mol

=> number of moles of S8 = 2.5g / 256.52 g/mol = 0.009746 mol

4) Limiting reactant

Theoretical ratio                           actual ratio

16 mol Fe / 3 mol S8                 0.0537 mol Fe / 0.009746 mol S8

5.33                                               5.50

So, there is a little bit more Fe than the theoretical needed to react all the S8, which means the S8 is the limiting reactant.

5) Calculate the number of moles of iron (III) produced with 2.5 g (0.009746 moles) of S8

3moles S8 / 8 moles Fe2S3 = 0.009746 moles S8 / x

=> x = 0.009746 * 8 / 3 moles Fe2S3 = 0.026 moles Fe2S3

6) Convert 0.026 moles Fe2S3 into grams

mass in grams = number of moles * molar mass

molar mass of Fe2S3 = 207.9 g/mol

mass = 0.026 mol * 207.9 g/mol = 5.40 g

7) Answer: option D)

3 0

2 years ago

Read 2 more answers

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

laiz [17]

<span>35.0 mL of 0.210 M
KOH molarity = moles/volume
find moles of OH do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+ they will cancel out: H+ + OH- -> H2O
but you'll have some left over,
pH=-log[H+] pOH
=-log[OH-] pH+pOH
=14</span>

3 0

3 years ago

Read 2 more answers