All categories

3 years ago

An aqueous solution is 40.0% by mass silver nitrate, AgNO3, and has a density of 1.47 g/mL.

Chemistry

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

Molarity = 3.46 M

Explanation:

Given data:

Percentage of silver nitrate = 40.0%

Density of AgNO₃ = 1.47 g/mL

Molarity of solution = ?

Solution:

40% means out of 100g we have 40 g of AgNO₃ and 60g solvent.

Number of moles of silver nitrate = mass /molar mass

Number of moles = 40 g/ 170 g/mol

Number of moles = 0.235 mol

Volume of solution:

Total mass of solution = 100 g

density = 1.47 g/mL which means there are 1.47 gram into 1 mL.

100 g/1 × 1mL/ 1.47 g × 1 L / 1000 mL

0.068 L

Molarity:

Molarity = number of moles / volume in L

Molarity = 0.235 mol / 0.068 L

Molarity = 3.46 M

You might be interested in

What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:

Valentin [98]

Answer : The correct expression will be:

$K=(K_1)^2\times K_2$

Explanation :

The chemical reactions are :

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)$ $K_1$

(2) $2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)$ $K_2$

The final chemical reaction is :

$N_2(g)+2O_2(g)\rightleftharpoons N_2O_4$ $K=?$

Now we have to calculate the value of $K$ for the final reaction.

Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

If the two equations are added then equilibrium constant will be multiplied.

Thus, the value of 'K' will be:

$K=(K_1)^2\times K_2$

6 0

3 years ago

In a nuclear reactor, when the control rods are lowered between the fuel rods, they slow the fission reactions.

rosijanka [135]

True when the boron control rods are lowered it slows the reaction

5 0

3 years ago

Read 2 more answers

Which substance loses electrons in a chemical reaction? the one that is oxidized, which is the oxidizing agent the one that is r

victus00 [196]

Answer:

The one that is oxidized or the reducing agent

Explanation:

Oxidation results in loss of electrons or an increase in oxidation state.

Reduction results in gain of electrons or decrease in oxidation state.

The element that undergoes oxidation is said to oxidised, similarly the element that undergoes reduction is said to be reduced.

In a redox reaction, both oxidation and reduction takes place.

If a substrate undergoes oxidation or is oxidised, it is also responsible for reduction of the other species, as the total number of electrons should always be conserved.

The substance that undergoes oxidation, releases some electrons, these electrons are taken by the other substrate and it undergoes reduction.

Hence the substance that undergoes oxidation is called the reducing agent, as it is responsible for reduction of other substrate, when oxidizing itself.

Similar thought works for a substance that undergoes reduction or is reduced, works as an oxidizing agent.

7 0

3 years ago

Read 2 more answers

BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.

Yakvenalex [24]

Answer:

$T_2=-125.58\°C$

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Whereas solving for the final temperature T2, we get:

$T_2=\frac{T_1P_2}{P_1}$

Thus, we plug in the given data (temperature in Kelvins) to obtain:

$T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C$

Best regards!

3 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

6 0

3 years ago