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zvonat [6]
4 years ago
5

Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation:

Chemistry
1 answer:
sergiy2304 [10]4 years ago
7 0
The balanced chemical reaction is:

<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.

2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3

4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3

63.8g N2H4  <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>
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What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2?
NemiM [27]

Hey there!:

Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :

MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)

Write the total ionic equation  for the reaction :

Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)

Therefore:

Cancel the spectator ions on both sides:

Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)


Hope that helps!

4 0
3 years ago
From where do the gems used in Jewlery come? A. they are mettallic ore B.they are nonmettallic silicates C. they are mettallic m
mixas84 [53]
The answer is D, because gems are usually not mettallic
6 0
3 years ago
Read 2 more answers
Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine
Lilit [14]

<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

<u>Explanation:</u>

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:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 34.9689 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 36.9659 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

  • Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.7577\times 100=75.77\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.7577)=0.2423\times 100=24.23\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

6 0
3 years ago
What is a lone pair?
gregori [183]

Answer:

I BELIEVE IT is c.

Explanation:

7 0
3 years ago
ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomp
mamaluj [8]

Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

4 0
3 years ago
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