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-BARSIC- [3]
2 years ago
15

A balloon inflated in a room at 25oC has a volume of 3.00 L. If the balloon is heated at a constant pressure, at what temperatur

e will the volume of the balloon double?
1. 596 K
2. 323 K
3. 298 K
4. 149 K
Chemistry
1 answer:
Alexeev081 [22]2 years ago
8 0

Answer:

596K

Explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 3.00 L

V2 = double of V1 = 2 × 3.00 = 6.00 L

T1 = 25°C = 25 + 273 = 298K

T2 = ?

Using V1/T1 = V2/T2

3/298 = 6/T2

Cross multiply

298 × 6 = 3 × T2

1788 = 3T2

T2 = 1788 ÷ 3

T2 = 596K

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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

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3 years ago
How many significant figures are in 4.800x10-3?​
seraphim [82]
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The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what
valina [46]

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

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3 0
3 years ago
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