The best definition of a proxy record is:

PLEASE HELP WITH ALL! Do the following Mole conversions:

ExtremeBDS [4]

Answer:

The answer to your question is below

Explanation:

1) number of molecules = x

23 g of CoCl₂

Molar mass CoCl₂ = 59 + 71 = 130 g

130 g ------------------ 6.023 x 10²³ molecules

23 g ----------------- x

x = (23 x 6.023 x 10²³) / 130

x = 1.066 x 10²³ molecules

2)

3.2 moles of H₃PO₄ to grams

Molar mass = 3 + 31 + (16 x 4)

= 98 g

98 g ------------------- 1 mol

x ------------------- 3.2 moles

x = (3.2 x 98)/1

x = 313.6 g

3)

2.2 moles of (NH₄)₂ to grams

Molar mass = (14 x 2) + (8 x 1)

= 36 g

36 g of (NH₄)₂ ---------------- 1 mol

x ---------------- 2.2 moles

x = (2.2 x 36)/1

x = 79.2 grams

4)

Molecules = ? 4.5 g of NH₄SO₂

Molar mass = 14 + 32 + 32 + 4

= 82 g

82 g of NH₄SO₂ --------------- 6.023 x 10²³ molecules

4.5 g ---------------- x

x = (4.5 x 6.023 x 10²³) / 82

x = 3.3 x 10²³ molecules

5)

Liters of N₂ = ? 7.9 x 10³⁰ particles

22.4 l --------------------- 6.023 x 10²³ particles

x -------------------- 7.9 x 10³⁰

x = (7.9 x 10³⁰ x 22.4) / 6.023 x 10²³

x = 2.93 x 10⁸ particles

7 0

3 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

$\frac{100}{56}$

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

$\frac{Actual\ yield}{Theoritical\ yield}\times 100$

$\frac{19.4}{25.71}\times 100$

= 75.44 %

6 0

3 years ago

What parts of a circuit is a conductor?

netineya [11]

Answer:

copper

Explanation:

These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load

3 0

3 years ago

Fire is combination of what? example ( water is a compound of hydrogen and oxygen)

nirvana33 [79]

Carbon dioxide, water vapour, oxygen and nitrogen

8 0

4 years ago

Consider the model of the nitrogen atom . Which electron configuration matches this model?

Galina-37 [17]

Answer:

The atomic number of Nitrogen is seven. So it contains seven protons and seven electrons in neutral form. Also, the electronic configuration cited above contains seven electrons among which two electrons are present in first shell and five electrons are present in valence shell respectively.

Explanation:

5 0

3 years ago