All categories

3 years ago

Does this sound good/ right? I hope so! Thank you! please give me criticism NICELY!!

Chemistry

2 answers:

jek_recluse [69]3 years ago

8 0

In my opinion you explained this perfectly and it also sounds correct but because you jumped from one topic to another and then back to the same one i think you should switch up your sentences. you should probably switch the second and last mainly because the second sentence doesn't mention borax like the rest do

lol sounds like im bossing yo around but you dont have to change it, its just something i would do. hope this helps!!!

Mamont248 [21]3 years ago

5 0

This sounds good so far.
May want to check how you say "and are spread way out" - a better way to say it would be "the crystals are spread out"

You might be interested in

Which one of the following compounds will be least soluble in water?

algol [13]

Water is a polar solvent. Since, 'Like dissolves like' a polar compound will be most soluble in water whereas a non-polar compound will be the least soluble.

The increasing order of polarity for the given organic compounds is:-

CH3OCH3 < CH3NH2 < CH3CHO < CH3CH2OH

Ether is the least polar and hence least soluble in water.

8 0

2 years ago

Read 2 more answers

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

What is the formula for number of moles?

kicyunya [14]

Answer:

number of moles=volume/molar volume or mass/molar mass

8 0

2 years ago

Read 2 more answers

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

A sample of methane gas has a volume of 355 mL at 25.0°C and 2414 mm Hg. If the temperature is raised to 100.°C and the pressure

Drupady [299]

Answer:

23019

Explanation:

5 0

2 years ago

Read 2 more answers