Who believed that a grain of sand could be divided indefinitely?

Does anyone know o I'm correct? If not what's the answer?

igomit [66]

I believe it means idea because a Web is out line to a a more structured idea

6 0

3 years ago

How many liters are there in 28g of B?

Juliette [100K]

Hi
The volume is 58.1
The moler mass is 10.8
Molar mass details
10.81B (1*10.81)

Hope this helps

5 0

3 years ago

7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.

beks73 [17]

Answer:

$\Delta H_{f,C_3H_4}=276.8kJ/mol$

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

$\Delta H_{rxn} =- m_wC_w\Delta T$

We plug in the mass of water, temperature change and specific heat to obtain:

$\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ$

Now, this enthalpy of reaction corresponds to the combustion of propyne:

$C_3H_4+4O_2\rightarrow 3CO_2+2H_2O$

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

$\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}$

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

$\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol$

Now, we solve for the enthalpy of formation of C3H4 as shown below:

$\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}$

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

$\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol$

Best regards!

7 0

3 years ago

The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28

ivanzaharov [21]

Answer: The mass of second isotope of indium is 114.904 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

.....(1)

Let the mass of isotope 2 of indium be 'x'

For isotope 1:

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

For isotope 2:

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

$114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu$

Hence, the mass of second isotope of indium is 114.904 amu

4 0

3 years ago

How did the construction of a parking lot, hiking trail, and playground affect the fish populations in the pond? PLEASE HELP.

Tpy6a [65]

A),b),c),d) answers?

4 0

3 years ago

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