Which wave characteristic determines color

Imagine that Ana has a block made of pure gold. If she cuts this block into two equal pieces, pick the statement that best descr

dimaraw [331]

Answer : The correct statement is:

The density of each piece is the same as that of the original block.

Explanation :

Intensive property : It is defined as a property of substance which does not change as the amount of substance changes.

Examples: Temperature, refractive index, density, hardness, etc.

According to question, if Ana has a block made of pure gold and she cuts this block into two equal pieces then the density of each piece is the same as that of the original block because density is an intensive property that does not changes until and unless material is changed.

That means density remains same as that of the original piece.

Hence, the correct statement is the density of each piece is the same as that of the original block.

6 0

3 years ago

Which cohesive forces are the strongest in water?

jeyben [28]

When cohesive force is stronger than the adhesive force: concave up meniscus, water forms droplets on surface

4 0

3 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

Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3 (taking into con

Misha Larkins [42]

Answer:

v = 1130 cm³

Explanation:

Given data:

Volume of sample = ?

Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)

Density of Al sample = 2.70 g/cm³

Solution:

Formula:

d = m/v

d = density

m = mass

v= volume

by putting values

2.70 g/cm³ = 3057g /v

v = 3057g /2.70 g/cm³

v = 1130 cm³

3 0

3 years ago

How many chlorine atoms are there in 12.2 g of ccl4? How many chlorine atoms are there in 12.2 of ? 4.77×1022 atoms?

mash [69]

Answer:- $1.91*10^2^3Cl$ atoms.

Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.

For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:

$12.2gCCl_4(\frac{1molCCl_4}{153.82gCCl_4})(\frac{4molCl}{1molCCl_4})(\frac{6.02*10^2^3Clatoms}{1molCl})$

= $1.91*10^2^3Cl$ atoms

So, there are $1.91*10^2^3Cl$ atoms in 12.2 grams of $CCl_4$ .

6 0

3 years ago

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