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4 years ago

Give the number of particles of each element in NH3, ammonia.

1 answer:

3 0

You'll need Avogadro's number for this. NH3 has 1 Nitrogen and 2 Hydrogen. You'll multiply the molar mass of each times Avogadro's number. So:

molar mass N x 6.022x10^23 x 1 Nitrogen= total A

Molar mass H x 6.022x10^23 x 2 Hydrogen= total B

Add total A+B = your answer

I don't have a calculator or periodic table but if you plug in the numbers you'll get your answer. Hope it helps.

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When potassium is added to water, what causes the explosion?

malfutka [58]

Answer:

The highly unstable pure sodium or potassium wants to lose an electron, and this splits the water atom, producing a negatively charged hydroxide ion and hydrogen and forming an explosive gas that ignites.

Explanation:

7 0

3 years ago

When 10.0 grams of CH4 reacts completely with 40.0 grams of O2 such that there are no reactants left over, 27.5 grams of carbon

BARSIC [14]

Answer:

$\boxed{\text{27.4 g CO$_{2}$; 22.5 g H$_{2}$O}}}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 16.04 32.00 44.01 18.02

CH₄ + 2O₂ → CO₂ + 2H₂O

m/g: 10.0 40.0

1. Moles of CH₄

$\text{Moles of CH}_{4} = \text{10.0 g CH}_{4} \times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}} = \text{0.6234 mol CH}_{4}$

2. Mass of CO₂

(i) Calculate the moles of CO₂

The molar ratio is (1 mol CO₂ /1 mol CH₄)

$\text{Moles of CO$_{2}$} = \text{0.6234 mol CH$_4$} \times \dfrac{\text{1 mol CO$_{2}$}} {\text{1 mol CH$_{4}$}} = \text{0.6234 mol CO$_{2}$}$

(ii) Calculate the mass of CO₂

$\text{Mass of CO$_{2}$} = \text{0.6234 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{27.4 g CO$_{2}$}\\\\\text{The mass of carbon dioxide formed is } \boxed{\textbf{27.4 g CO$_{2}$}}$

3. Mass of H₂O

(i) Calculate the moles of H₂O

The molar ratio is (2 mol H₂O /1 mol CH₄)

$\text{Moles of H$_{2}$O}= \text{0.6234 mol CH}_{4} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{1 mol CH$_{4}$}} = \text{1.247 mol H$_{2}$O}$

(ii) Calculate the mass of H₂O

$\text{Mass of H$_{2}$O} = \text{1.247 mol H$_{2}$O } \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{22.5 g H$_{2}$O}\\\\\text{The mass of water formed is } \boxed{\textbf{22.5 g H$_{2}$O}}$

4 0

3 years ago

What can be concluded about the atom from knowing that oxygen 18 has an atomic number of 8?

Vinvika [58]

<span>The atomic number of a neutral atom is equal to the number of protons and the number of electrons of the atom. The atomic weight meanwhile is equal to the sum of the number of protons and number of neutrons.Hence we can say that the number of protons is 8 and the number of neutrons is 10. </span>

7 0

3 years ago

Read 2 more answers

Using the equation below, calculate the number of grams of H2O that are formed from the reaction of 34.0 g of NH3.

Mumz [18]

Answer:

54 grams of H₂O

Explanation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Ratio is 4:6.

Let's convert the mass of ammonia in moles

Mass / Molar mass = Mol

34 g / 17 g/m = 2 moles

If 4 moles of ammonia are needed to produce 6 mol of water

2 mol of NH₃ will produce ( 2 .6 )/4 = 3 moles

Molar mass water = 18 g/m

Mass = mol . molar mass → 3 m . 18 g/m = 54 grams

8 0

3 years ago

How much heat is released when 10.0 grams of a substance cools 15 degrees? The specific

VLD [36.1K]

Answer:

Q = 525 J

Explanation:

Given that,

Mass, m = 10 grams

The change in temperature, $\Delta T=15^{\circ} C$

The specific heat of the substance is 3.5 J/g °C

We need to find the amount of heat released in the process. It can be given by the formula as follow :

$Q=mc\Delta T\\\\Q=10\ g\times 3.5\ J/g^{\circ} C\times 15^{\circ} C\\Q=525\ J$

Hence, 525 J of heat is released.

4 0

3 years ago