Question given above

Is density defined as the volume in a given mass orthe mass in a given volume?

wariber [46]

Answer:

mass in a given volume

Explanation:

Density is the mass in a given volume or mass per unit volume of a substance.

Mass is defined by the amount of matter present in a substance.

Volume is the space occupied by a body.

Density is the mass divided by the volume

Mathematically;

Density = $\frac{mass}{volume}$

3 0

3 years ago

This graph illustrates that, under conditions of constant pressure, as the temperature of liquid water increases, its volume ___

Lunna [17]

it varies.........................

8 0

3 years ago

Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso

Sauron [17]

Answer:

For A: The number of arsenic atoms are $3.4\times 10^{21}$

For B: The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Explanation:

For A:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of dimercaprol = 696 mg = 0.696 g (Conversion factor: 1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

$\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.0056 moles of dimercaprol will contain $0.0056\times 6.022\times 10^{23}=3.4\times 10^{21}$ number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, $3.4\times 10^{21}$ number of dimercaprol molecules will bind with = $1\times 3.4\times 10^{21}=3.4\times 10^{21}$ number of arsenic atoms

Hence, the number of arsenic atoms are $3.4\times 10^{21}$

For B:

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

$\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100$ ......(1)

For mercury:

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

$\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%$

For thallium:

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

$\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%$

For chromium:

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

$\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%$

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0

3 years ago

Calculate the amount of formula units in 2.10 moles of MgCl₂

Eva8 [605]

MgqI% = the formula and mass for H20 and that should be your answer for water

5 0

3 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is $CO3^{-2}$ , and the ion nitrate is $NO3^{-}$ .

Sodium is in group 1, so it must lose one electron to be stable, and be the cation $Na^{+}$ . Silver has only one electron too, so the cation will be $Ag^{+}$ .

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0

3 years ago