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AysviL [449]
3 years ago
13

Caffeine is used commercially as a(n):

Chemistry
1 answer:
Nataly [62]3 years ago
8 0
I guess it would be a psychoactive substance i'm not really sure.. but i hope this helped 
You might be interested in
How many grams are in 3.4 moles of C2H5OH molecules? (use molar mass and round answer to the correct number of sig figs)
Salsk061 [2.6K]
Molar mads of C2H5OH:
12*2+1*5+16+1
24+5+17
46
no of moles = mass in grams/molar mass
3.4=m/46
3.4*46=m
m=156.4
m=156g
5 0
3 years ago
^^^ AHH HELPOP ASAPP
Liula [17]
The apparent effects of the gravity of unseen matter on visible objects
3 0
3 years ago
Ans with solution...
Licemer1 [7]

Answer:

The answer to your question is:

Vol of NO2 = 11.19 L

Vol of O2 = 2.8 L

Explanation:

Data

N2O5 = 56 g

STP     T = 0°C = 273°K

           P = 1 atm

MW N2O5 = 216 g

Gases law = PV = nRT

Process

                   216 g of N2O5 ---------------- 1 mol

                     54 g               -----------------  x

                    x = (54 x 1) / 216

                    x = 0.25 mol of N2O5

                   2 mol of N2O5 -----------------  4 mol of NO2

                   0.25 mol          ------------------    x

                   x = (0.25 x 4) / 2 = 0.5 mol of NO2

                   V = nRT/P

                   V = (0.5)(0.082)(273) / 1 = 11.19 L

                   2 mol of N2O5 ----------------- 1 O2

                   0.25 N2O5 ----------------------  x

                   x = (0.25 x 1) / 2 = 0.125 mol

                   Vol = (0.125)((0.082)(273) / 1 = 2.8 L

3 0
4 years ago
Calculate the morality of each of the following solutions: a. 15.4 g KCl in 289.2 mL solution b. 14.4 g of CaCl2 in 0.614 L solu
Nesterboy [21]
The answer is:
a. 0.712 M
b. 0.210 M
c. 0.336 M

Molarity is a measure of the concentration of solute in a solution.
It can be expressed as moles of solute ÷ volume of solution:
c = n ÷V
where:
c - concentration of solute,
n - moles of solute
V - volume of solution

n can be expressed as:
<span>n = m ÷ Mr
</span>where:
<span>n - moles of solute
</span>m - mass of solute
Mr - relative molecular mass

a. We know volume:
V = 289.2 mL = 0.2892 L
We need n and c.

n = m ÷ Mr
m = 15.4 g
Mr (<span>KCl) = 74.55 g/mol
n = </span>15.4 g ÷ <span>74.55 g/mol
n = 0.206 mol</span>

Thus, 
c = 0.206 mol ÷ <span>0.2892 L
c = 0.712 mol/L = 0.712 M

</span>b. We know volume:
V = 0.614 L
We need n and c.

n = m ÷ Mr
m = 14.4 g
Mr (CaCl₂<span>) = 110.98 g/mol
n = </span>14.4 g ÷ <span>110.98 g/mol
n = 0.129 mol</span>

Thus, 
c = 0.129 mol ÷ <span>0.614 L
c = 0.210 mol/L = 0.210 M
</span>
c. We can use formula:
m₁V₁ = m₂V₂
<span>m₁ = 3 M
</span><span>V₁ = 28 mL= 0.028 L
</span><span>m₂ = ?
</span><span>V₂ = 0.250 L
</span>Thus:
3 M × 0.028 L = m₂× <span>0.250 L
</span> m₂ = 0.336 M
4 0
3 years ago
During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul
gregori [183]

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

7 0
3 years ago
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