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3 years ago

At terminal velocity, which two forces are balanced? *

Chemistry

1 answer:

OlgaM077 [116]3 years ago

8 0

Answer:

friction and air resistance

at terminal velocity, the weight of the object due to gravity is balanced by the frictional forces, and the resultant force is zero.

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Who's theory states that atoms are made up of empty space with electrons around a positively charged mass?

emmasim [6.3K]

That would be Rutherford's Theory.

5 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

4 years ago

A container is filled with helium gas. It has a volume of 2.25 liters and contains 9.00 moles of helium. How many moles of heliu

Darina [25.2K]

Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

Explanation:

Given: $V_{1}$ = 2.25 L, $n_{1}$ = 9.0 mol

$V_{2}$ = 1.85 L, $n_{2}$ = ?

Formula used to calculate the moles of helium are as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{2.25 L}{9.0 mol} = \frac{1.85 L}{n_{2}}\\n_{2} = \frac{1.85 L \times 9.0 mol}{2.25 L}\\= 7.4 mol$

Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

7 0

3 years ago

f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration

goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0

3 years ago

HELP ASAP

Bezzdna [24]

A. I think sorry if it’s wrong

5 0

3 years ago