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disa [49]
2 years ago
14

The distance from the sun to the earth is 1.50 x 10^8 km. The speed of light is 3.00 x10 ^8m/s. How many round trips between Ear

th and the sun could a beam of light make in one day?
Chemistry
1 answer:
icang [17]2 years ago
6 0

Answer:

240*60/8=180

Explanation:

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A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o
Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

c=\frac{n}{V}

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol

Now, the total volume of this solution can be found by adding the volume values of each component:

V_total=46.2 mL+80.5 mL=126.7 mL

Finally, dividing the moles found by the total volume will yield the final molarity:

c_{final}=\frac{138.62 mmol}{126.7 mL}= 1.09 M

6 0
3 years ago
Helppppp pls it’s due by today
Andrews [41]

Answer:

i can't see the picture

Explanation:

4 0
3 years ago
Read 2 more answers
S + 6 HNO3 H2SO4 + 6 NO2 + 2 H2O
exis [7]

Answer:

              101.50 g H₂O

Explanation:

The mole ratio of HNO₃ and H₂O is 6 : 2

Hence, 16.9 moles of HNO₃ will produce = 2/6×16.9 = 5.63 moles of H₂O

Also,

Mass = Moles × M.Mass

Mass = 5.63 mol × 18.02 g/mol

Mass = 101.50 g H₂O

5 0
2 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
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What does the release or absorption of energy indicate
Illusion [34]
Chemical change
Hope this helps!
4 0
3 years ago
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