All categories

2 years ago

Reacting 35.4 ml of 0.220 m agno3 with 52.0 ml of 0.420 m k2cro4 results in what mass of solid formed

1 answer:

7 0

Answer is: 1.29 grams of solid formed.
Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.

You might be interested in

What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)

amid [387]

Answer:

$m=59.3gCO_2$

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

$M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol$

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

$mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol$

Finally, the mass by using the molar mass:

$m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2$

Best regards.

4 0

2 years ago

What is the difference between atomic mass, relative atomic mass and average atomic mass

vesna_86 [32]

$\bold{\huge{\underline{\purple{ Answer }}}}$

Difference between Atomic mass, relative atomic mass and average atomic mass :-

<h3>Atomic Mass :-</h3>

Atomic mass is the mass of neutrons and protons present in the nucleus of an atom .
It is always calculated for a single element and having direct value
For isotopes also, the atomic mass is calculated separately . Example :- Carbon 12 , carbon 13 and carbon 14 have different atomic mass. 
The SI unit of Atomic mass is " u" and "amu"

<h3>Relative Atomic mass :-</h3>

Relative atomic mass is mean mass of the atoms of an element which is compared to the 1/12th mass of carbon - 12 .
Carbon - 12 is taken as a relative when we calculate the relative atomic mass of any element
For calculating relative atomic mass, we need to know the masses, percentage and abundance of all types of elements
Relative atomic mass is a dimension less quantity

<h3>Average Atomic Mass :-</h3>

Average atomic mass is the average mass of an atoms of a particular element by considering it's isotopes
While we calculate average atomic mass is a standardized number. Whereas, Average atomic mass sometimes varies geologically .
It also includes percentage, abundance and masses of given element .
In average atomic mass, We do not compare mean value with the 1/12 mass of carbon - 12
The unit of Average atomic mass is "Amu" or " u " .

5 0

1 year ago

46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride

slega [8]

Answer:

$m_{HgCl_2}=42.7gHgCl_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4$

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

$m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2$

Best regards.

3 0

2 years ago

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150

Usimov [2.4K]

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

$M=\frac{Moles of substance X}{Volume of the solution}$

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

$11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}$

Then you can solve for the volume of the solution:

$Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL$

Hope it helps!

3 0

3 years ago

What is atom?????????????

kiruha [24]

Answer:

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. ... The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons. A proton is positively charged.

6 0

1 year ago

Read 2 more answers