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Margaret [11]
3 years ago
8

How many grams of glucose, C6H12O6, in 2.47 mole?

Chemistry
2 answers:
konstantin123 [22]3 years ago
7 0
Answer:
number of grams = 444.6 grams

Explanation:
From the periodic table, we can find that:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
This means that:
molar mass of C6H12O6 = 6(12) + 12(1) + 6(16) = 180 grams

Now, number of moles can be calculated using the following rule:
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass = 2.47 * 180
mass = 444.6 grams

Hope this helps :)
statuscvo [17]3 years ago
6 0
First, we need to find the atomic mass of C_{6}H_{12}O_{6}.

According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16

As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of  C_{6}H_{12}O_{6} = 6 * 12.01 + 12 * 1.008 + 6 * 16

The molar mass of  C_{6}H_{12}O_{6} = 180.156 grams/mole

Now that we have the molar mass of  C_{6}H_{12}O_{6}, we can find the grams of glucose by using:

mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)

Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams

Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams

-i
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Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
Find the volume of 20g of H₂ at STP<br><br>​
ANEK [815]

Answer:224

Explanation:

We should answer it with Stoichiometry

We say: 20 g H2× (1 mol/ 2g)× ( 22.4 lit/ 1 mol) = 224

Means: we have 20 grams and every 2g H2, equals to 1 mol of it and every 1 mol of H2, equals to 22.4 lit( because of STP)

hope you got this:)

6 0
3 years ago
Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit
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The dissociation equation will be

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Initial                 0.006                        0                          0

Change         -0.006 X 0.053        +0.006 X 0.053      -0.006 X 0.053

Equlibrium     0.006 -0.006 X 0.053      0.006 X 0.053    0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

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3 years ago
4. An object's resistance to any change in its motion is the ___ of the object. (1 point)
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