All categories

3 years ago

Identify the sentence that correctly uses an ellipsis.

Chemistry

2 answers:

zloy xaker [14]3 years ago

5 0

C)"his refusal to cooperate with the police and name the culprit...resulted in his arrest..."

34kurt3 years ago

3 0

B) “His refusal to cooperate with the police and name the culprit...resulted in his arrest.”

You might be interested in

Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3 (taking into con

Misha Larkins [42]

Answer:

v = 1130 cm³

Explanation:

Given data:

Volume of sample = ?

Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)

Density of Al sample = 2.70 g/cm³

Solution:

Formula:

d = m/v

d = density

m = mass

v= volume

by putting values

2.70 g/cm³ = 3057g /v

v = 3057g /2.70 g/cm³

v = 1130 cm³

3 0

3 years ago

Using GRIGNARDS REAGENT convert methane to ethanol

slamgirl [31]

Answer:

Explanation:

6 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$