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mina [271]
3 years ago
14

199.5 grams unrefined dark crystalline sugar to cups

Chemistry
1 answer:
Dmitry [639]3 years ago
4 0

Answer:

0.9975 cup  

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

 199.5 g brown sugar = 199.5× 1/200 .0

 199.5 g brown sugar = 0.9975 cup

A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.

Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.

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∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821
Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

<h3>4) 0.26 atm </h3><h3 />
6 0
3 years ago
(sand) and (Sand with water) both of them are heterogeneous<br> mixture isn't it?​
patriot [66]

Answer:

yes true

Explanation:

both are heterogeneous

8 0
3 years ago
In the laboratory you are asked to make a 0.303 m cobalt(II) sulfate solution using 275 grams of water. How many grams of cobalt
valentinak56 [21]

Answer: 12.92g of CoSO4

Explanation:

Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol

Molarity of CoSO4 = 0.303mol/L

Mass conc. In g/L = Molarity x molar Mass

= 0.303x155 = 46.965g/L

275 grams of water = 0.275L of water

46.965g of CoSO4 dissolves in 1L

Therefore Xg of CoSO4 will dissolve in 0.275L i.e

Xg of CoSO4 = 46.965x0.275 = 12.92g

Therefore 12.92g of CoSO4 is needed

4 0
3 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
3 years ago
Plssssssssssssssssssmejepleplele
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Answer:

Yes the two of the answer is True

3 0
3 years ago
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