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2 years ago

Carbon dating requires that the object being tested contain

Chemistry

1 answer:

luda_lava [24]2 years ago

4 0

Answer:

Organic Material

Explanation:

Carbon Dating is the process in which the age of a piece of organic matter is determined by the proportions of carbon isotopes it contains.

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3 years ago

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2 years ago

How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?

slava [35]

Answer:

A. 0.90 L.

Explanation:

NaOH solution will react with H₂SO₄ according to the balanced reaction:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.

For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

∴ (MV) NaOH = (xMV) H₂SO₄.

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH = 2(0.6 L)(3.0 M)/(4.0 M) = 0.90 L.

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2 years ago

Ductile means...

balu736 [363]

Answer:

Explanation:

3 0

2 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago