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Irina18 [472]
3 years ago
5

A baker is decorating cupcakes using 3 cherries for every 1 cupcake. If the baker has 12 cherries and 5 cupcakes, what is the th

eoretical yield?
Chemistry
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

The baker would only be able to make 4 cupcakes each having 3 cherries on it.

Explanation:

Since the focus is on cherries you can ignore the cupcakes. So if you have 12 cherries and 3 PER cupcake you divide. 12 by 3 to get 4. Even though there will be one cupcake left, they could always give it to me! lol

Hope this helps and have a great day!

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A certain compound has the percent composition (by mass) 85.63% C and 14.37% H. The molar mass of the compound is 42.0 g/mol. Ca
AlekseyPX

Answer:

The molecular formula is C3H6

Explanation:

Step 1: Data given

Suppose the compound has a mass of 100 grams

The compound contains:

85.63 % C = 85.63 grams C

14.37 % H = 14.37 grams H

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles

Moles = grams / molar mass

Moles C = 85.63 grams / 12.01 g/mol

Moles C = 7.130 moles

Moles H = 14.37 grams / 1.01 g/mol

Moles H = 14.2 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 7.130 moles / 7.130 moles = 1

H = 14.2 moles / 7.130 moles = 2

The empirical formula is CH2

The molar mass of CH2 = 14 g/mol

Step 4: Calculate molecular formula

We have to multiply the empirical formula by n

n = 42 / 14 = 3

n*(CH2) = C3H6

The molecular formula is C3H6

8 0
3 years ago
Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy
gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol    

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure ,  Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$  ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

  $Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

    $2P=PE^{6.4 \times 10^{-4}P}$

   $2=E^{6.4 \times 10^{-4}P}$

  $\ln 2 = 6.4 \times 10^{-4}P$

  ∴  P = 1083.04 atm

c). dG = Vdp -S dt  at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $

            $=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

           $=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol  

7 0
3 years ago
In which substance does clorine have an oxidation number of +5?​
Oksi-84 [34.3K]

Answer:oxidation number H = +1

oxidation number O = -2

let x = oxidation number Cl

in HClO3

+1 + x -6 =0

x = +5

Explanation:

3 0
4 years ago
What kind of interaction is not involved in the binding of a substrate to a normally functioning enzyme?
matrenka [14]
It seems that you have missed the necessary options for us to answer this question, so I had to look for it. Anyway, here is the answer. The kind of interaction that is NOT involved in the binding of a substrate to a normally functioning enzyme is a<span> permanent covalent bond. Hope this helps.</span>
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3 years ago
A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of
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<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0
3 years ago
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