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2 years ago

How can an object enter Earth's orbit?

1 answer:

3 0

When an object enters the Earth's atmosphere, it experiences a few forces, including gravity and drag. Gravity will naturally pull an object back to earth.

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The radius of a hydrogen atom is 37 pm and its mass is 1.67 × 10-24 g. What is the density of the atom in grams per cubic centim

RoseWind [281]

Answer:

V = 4/3 * 3.1416 * (37x10-10)3

V = 2.12x10-25 cm3

d = m/V

d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3

The difference in temperature, let's convert F to ºC:

ºC = -80-32/1.8 = -62.22 ºC

dT = -92.6 + 62.2 = -30.4 ºC

4 0

3 years ago

Read 2 more answers

If 75 grams of oxygen react, how many grams of aluminum are required?

german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

O₂ : Al

3 : 4

2.34 : 4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0

2 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

3 years ago

What is the pH for a 0.10 M HCI solution at 25<br> degreesCelsius?

11Alexandr11 [23.1K]

Answer:

1.0

Explanation:

Hydrochloric acid is a strong acid, that is, an acid that dissociates completely, according to the following reaction.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

Then, the concentration of H⁺ will be equal to the initial concentration of the acid, i.e., 0.10 M.

We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 0.10 = 1.0

3 0

3 years ago

At constant force, acceleration varies (directly, inversely) with mass. When subjected to the same amount of net external force,

lora16 [44]

Answer:

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7 0

2 years ago

Read 2 more answers

How can an object enter Earth's orbit?​

How can an object enter Earth's orbit?