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11111nata11111 [884]
3 years ago
14

Calculate the number of sucrose molecules in a 75.0 gram sample.

Chemistry
1 answer:
zzz [600]3 years ago
5 0

Answer:

1.32*10^23 molecules

Explanation:

sucrose formula: C12H22O11

molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol

75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11

0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)

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The soda an air is most likely bubbling up and is going to explode.
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Identify at least three control variables that could be used with this investigation: What temperature allows for bacterial grow
Over [174]

Answer:

Time allowed for incubation, size of the dish, the amount of light, amount of agar, the type of agar…etc

Explanation:

6 0
3 years ago
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A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen. After cooling to room temperature, thr
Tpy6a [65]
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
5 0
3 years ago
A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0
3 years ago
How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05<br> g/mol.
dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

n = 2.45/6.02 × 10^(24-23)

n = 0.407 × 10¹

n = 4.07moles

Using mole = mass/molar mass

Where; molar mass of Fp3Bz2. is 97.05

g/mol.

mass = molar mass × mole

mass = 97.05 × 4.07

mass = 394.99g

4 0
3 years ago
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