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steposvetlana [31]
2 years ago
13

Fill in the blanks

Chemistry
1 answer:
daser333 [38]2 years ago
6 0

Answer:

Here is the answers I got 8 out of 8 correct :D))

Explanation:

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Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction
alexdok [17]

Answer:

see explanation

Explanation:

Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction occurs, simply write only NR.

Ag (+1) + NO3(-1) + 2 Na(+1) + Co3 (-2)--> Ag2CO3 (s) + 2 Na (+1) + 2NO3(-1)

8 0
2 years ago
A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centr
IRISSAK [1]
<h3><u>Answer;</u></h3>

<em><u> = 48,828.125 mi/hr²</u></em>

<h3><u>Explanation and solution</u>;</h3>
  • <em><u>Centripetal acceleration is the rate of change of angular velocity. Centripetal acceleration occurs towards the center of the circular path along the radius of the circular path</u></em>.
  • Centripetal acceleration is given by; <em>V²/r ; </em>

<em>V = 125 mi/h and r = 0.320 miles </em>

  • <em>Thus; centripetal acceleration = 125²/0.320 </em>

                                                  =15625/0.320

                                                 <em><u> = 48,828.125 mi/hr²</u></em>

4 0
2 years ago
Read 2 more answers
"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi
ycow [4]

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

7 0
3 years ago
What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S
Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat,  the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • <u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c= Heat Capacity of Liquid= 4.184 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 \frac{J}{gC}× 185.5 g× (- 25.6 °C)

Solving:

<u><em>Q1= -19,868.98 J</em></u>

  • <u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× \frac{1mol}{18 grams}= 10.30 moles, where 18 \frac{g}{mol} is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 \frac{kJ}{mol}

Replacing:

Q2= 10.30 moles×6.01 \frac{kJ}{mol}

Solving:

<u><em>Q2=61.903 kJ= 61,903 J</em></u>

  • <u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

  • c = Heat Capacity of Solid = 2.092 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 \frac{J}{gC} × 185.5 g× (-10.70) °C

Solving:

<u><em>Q3= -4,152.3062 J</em></u>

<h3>Total heat required</h3>

The total heat required is calculated as:  

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

Learn more about calorimetry:

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brainly.com/question/23578297?referrer=searchResults

7 0
2 years ago
Are the following combinations allowed? If not, show two ways to correct them:
mafiozo [28]

The following combination of n=3 ; l=1 ; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

<h3>Is the following combination n=3; l=1; ml=-2 allowed or not.? If not, suggest two ways through which it can be corrected.</h3>

The following combination of n=3 ; l=1 ; ml=-2 is not allowed.

There are several rules that need to be followed for assigning electron quantum numbers. They are:

1. Principal quantum number should be 1 ≤ n

2. Azimuthal quantum number, 0 ≤ l ≤ n − 1

3. Magnetic quantum number, -l ≤ ml ≤ l

4. Spin quantum number as either -1/2 or +1/2

For n = 3,

l should be n - 1 or n - 2 or n - 3 = 2, 1, 0 respectively.

If we choose l = 1 then ml should be -1, 0 and +1

Therefore, one way to correct the combination would be to change the magnetic quantum number to -1

If we choose l = 2 then ml would be -2, -1, 0, +1, +2

Thus, another way to correct the combination is to choose the azimuthal quantum number as 2.

Thus, the following combination of n=3; l=1; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

To learn more about quantum numbers refer:

brainly.com/question/5927165

#SPJ4

5 0
1 year ago
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