What is the purpose of this material in the nucleus?*

4. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0˚C in a calorimeter,

myrzilka [38]

Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

Volume of 1.00 M Na2SO4 = 1.00 L

enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL

Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

Q= m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

7 0

3 years ago

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

shutvik [7]

Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.

3 0

3 years ago

Radar detectors use radio waves to measure the speed of moving

bagirrra123 [75]

Answer:

1.7x10^8 Hz

Explanation:

Frequency could be explained as the number of occurrences of a repeating event at a time

Given:

wavelength = 1.8 meters

The frequency f of the waves can be calculated using f = c / λ

Where c (m/s) is the speed of the wave

λ (m) is the wavelength

Speed c= 3*10^8 m/s

Frequency f= 3*10^8 /1.8

Frequency= 1.7x10^8 Hz

Therefore,the frequency of waves from a radar detector is 1.7x10^8 Hz

6 0

3 years ago

Why are the oxidation and reduction half-reactions separated in an<br> electrochemical cell?

Aleksandr [31]

Answer:

<h2>It makes the current viable enough to pass through an exterior wire.</h2>

Explanation:

Electrochemical cells primarily comprise of two half-cells. These half-cells assist in isolating the oxidation and reduction half-reactions. These two reactions are linked by a wire which allows the current to move from one edge to the other. The oxidation at the anode and the reduction take place at the cathode and the addition of a salt bridge helps in completing the circuit and permits the current to flow and leads to the generation of electricity.

7 0

3 years ago

A 40.0-L sample of fluorine is heated from 90.0°C to 186.0°C. What volume will the sample occupy at the higher temperature?

jasenka [17]

Answer:

The volume will be 82.67 L

Explanation:

Charles's Law is the relationship between the volume and temperature of a certain amount of ideal gas. In this way, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} = k$

Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1} =\frac{V2}{T2}$

In this case, you know:

V1= 40 L
T1= 90 °C
V2= ?
T2= 186 °C

Replacing:

$\frac{40L}{90C} =\frac{V2}{186 C}$

Solving:

$V2=\frac{40L}{90C} *186 C$

V2= 82.67 L

<u><em>The volume will be 82.67 L</em></u>

5 0

3 years ago