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8_murik_8 [283]
2 years ago
5

Help plisssssssssss I need helppp

Chemistry
1 answer:
Makovka662 [10]2 years ago
6 0

Answer:

Explanation:

you would have to look more around the page, for example look at some ways that you can right down.

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How many atoms are there in 15.0 grams of Sulfur (S)?
aev [14]

Answer:

C

Explanation:

Mol x Avogadro's number

Mol : mass / molar mass

Avogadro's no: 6.02E+23

6 0
3 years ago
Which metal is heaviest ?
const2013 [10]
Titanium is the haviest metal .
8 0
3 years ago
Read 2 more answers
If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol
igomit [66]

Answer:

\Delta _RH=4x10^{-3}\frac{kJ}{mol}

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}

Best regards.

5 0
2 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
3 years ago
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