Givin 11 sodium 23, what is the atomic mass

What is 450 g of flour a measure of?

Marat540 [252]

The right answer is D - mass.

3 0

3 years ago

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LIMITING REACTANT!! Please help I’m very confused.

alexira [117]

Answer:

We'll have 1 mol Al2O3 and 3 moles H2

Explanation:

Step 1: data given

Numer of moles of aluminium = 2 moles

Number of moles of H2O = 6 moles

Step 2: The balanced equation

2Al + 3H2O → Al2O3 + 3H2

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

Aluminium is the limiting reactant. It will completely be consumed (2 moles).

H2O is in excess. There will react 3/2 * 2 = 3 moles

There will remain 6 - 3 = 3 moles

Step 4: Calculate moles products

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

For 2 moles Al we'll have 2/1 = 1 mol Al2O3

For 2 moles Al We'll have 3/2 * 2 = 3 moles H2

We'll have 1 mol Al2O3 and 3 moles H2

8 0

3 years ago

What kind of bond holds together atoms within a molecule?

sesenic [268]

I think it would be these three answers ionic , covalent , and polar covalent

7 0

3 years ago

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Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.

Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

To find the number of moles:

number of moles = $\frac{number of particles}{Avogadro's number}$

number of moles = $\frac{6.29 x 10^{23} }{6.02 x 10^{23} }$

number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

#learnwithBrainly

7 0

3 years ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.