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3 years ago

If Earth is in the portion of its orbital path that is further from the Sun, then what can be determined about its rate

Chemistry

2 answers:

Readme [11.4K]3 years ago

6 0

Answer:

The correct answer is "Its rate will be slower than when it is closer to the Sun".

Explanation:

The Earth passes through its furthest point from the Sun in its orbit, which is known as aphelion. According to Kepler's second law, the greater the distance, the lower the orbital velocity of translation. This happens because in its displacement around the Sun, the Earth does not have a perfect circular trajectory but an elliptical one. Kepler stated that the line connecting the planets and the Sun covers the same area in the same amount of time, so that when the planets are close to the Sun in its orbit, they move faster than when they are farther away.

Have a nice day!

Maurinko [17]3 years ago

4 0

If the earth's orbit is far from the sun, then, its rate will be slower than when it is closer to the Sun. When gravitational field lines get closer together, the magnetic force is strong. We know that the heavier the body is, the stronger its gravitational pull.

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Consider the elements in the periodic table. The stair-step line between the pink squares and the yellow squares separates the _

photoshop1234 [79]

The choices for this are as follows:

A) gases; solids
B) metals; nonmetals
C) nonmetals; metals
D) reactive; nonreactive

I think the correct answer is option B. The stair-step line between the pink squares and the yellow squares separates the metals from the nonmetals. Hope this helps.

6 0

3 years ago

What is the mass of solid NH4Cl formed when 75.5g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas remai

taurus [48]

Answer:The volume of the remaining gas that is ammonia is 23.85 L.

Explanation:

$NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)$

Moles of $NH_3=\frac{\text{mass of} NH_3}{\text{Molar mass of}NH_3}=\frac{75.5 g}{17.03 g/mol}=4.43 mole$

Moles of HCl of gas = $\frac{\text{mass of} HCl}{\text{Molar mass of}HCl}=\frac{75.5 g}{36.5 g/mol}=2.06 mol$

According to reaction 1 mole of HCl reacts with 1 mol of $NH_3$ then 2.06 moles of HCl will react with = 2.06 moles of $NH_3$

Moles left of ammonia left = 4.43 - 2.06 = 2.36 moles

Volume of the gas will be given by Ideal gas equation: PV=nRT

Pressure = 752 mmHg = 752 × 0.0031 atm = 2.33 atm

R = 0.08026 L atm/K mol

V = ? , n = number of moles of ammonia

Temperature = 14 °C = 14 + 273 K = 287 K(0°C = 273K)

$V=\frac{2.36 mol\times 0.08206 L atm/K mol\times 287 K}{2.33 atm}=23.85 L$

The volume of the remaining gas that is ammonia is 23.85 L.

8 0

3 years ago

Read 2 more answers

Why is it important to assume that the rate of radioactive decay has remained constant over time?

alexira [117]

It remains constant because it is an element. Just like Chernobyl, there is still active radiation. Radiation is something that stays in the air for a verrrry long time. Radioactivity kill cells, so lets say you have your phone on but its in your pocket.the radiation from the phone is killing your cells.

4 0

3 years ago

Question 9 (2 points)

KonstantinChe [14]

Answer:

0.39 moles

Explanation:

To find how many moles are in 50.0 g of CaC₂O₄ you divide the grams of the sample by the molar mass of the compound;

$\frac{50.0 g}{128.097 g/mol}$ =0.39 mol

The grams cancel out and you are left with moles!

I hope this help ^-^

7 0

3 years ago

Read 2 more answers

Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.60ft.

garik1379 [7]

Given:

Dimensions of the room= 12 ft * 15 ft * 8.60 ft

To determine:

The amount of HCN that gives the lethal dose in the room with the given dimensions

Explanation:

As per the World Health Organization, the lethal dose of HCN is around 300 ppm

300 ppm = 300 mg of HCN/ kg of inhaled air

Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³

Now, 1 ft³ = 28316.8 cm³

Therefore, the calculated volume of air corresponds to:

1548 * 28316.8 = 4.383 * 10⁷ cm3

Density of air (at room temperature 25 C) = 0.00118 g/cm3

Thus mass of air corresponding to the calculated volume is

Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

= 5.172*10⁴ g = 51.72 kg

Lethal amount of HCN corresponding to 51.72 kg of air would be.

= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg

Ans: Lethal dose of HCN = 15.5 g

7 0

3 years ago