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3 years ago

How many atoms are in a 591 g sample of gold?

Chemistry

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

You use the fact that 1 mole of any substance contains exactly 6.022⋅1023 atoms or molecules of that substance - this is known as Avogadro's number. In your case, 1 mole of gold will have exactly 6.022⋅1023 atoms of gold.

So it would be answer 3690 atoms or 3.60x1025

Explanation:

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2. ATO kg bouting ball would require what force to accelerate down an alleyway at a rate of ons?

Mars2501 [29]

Answer:

2. 30N

3. 5,000N

4. 15 kg

5. 2,800 kg

Explanation:

2. A 10 kg bowling ball would require what force to accelerate down an alleyway at a rate of 3m/s² ?

Notice that I completed the question with the garbled and missing values:

Data:

F = ?

m = 10 kg

a = 3m/s²

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=10kg\times 3m/s^2=30N$

3. Salty has a car that accelerates at 5 m/s². If the car has a mass of 1000 kg, how much force does the car produce?

Notice that I arranged the typos.

Data:

F = ?

m = ?

a = ?

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=1,000kg\times5m/s^2=5,000N$

4. What is the mass of a falling rock if it produces a force of 147 N?

Data:

F = 147N

m = ?

a = falling rock

Physical principles:

neglecting air resistance ⇒ a = g: gravitational acceleration: 9.8m/s²

Newton's second law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 147 N and a = g = 9.8m/s², and compute

$m=\dfrac{147N}{9.8m/s^2}=15Kg$

5. What is the mass of a truck if it produces a force of 14,000 N while accelerating at a rate of 5 m/s²?

Data:

F= 14,000N

m = ?

a = 5m/s²

Physical principles:

Second Newton's law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 14,000 N and a = 5m/s², and compute

$m=\dfrac{14,000N}{5m/s^2}=2,800kg$

6 0

3 years ago

An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 36 mg every 30. min . The IV b

tigry1 [53]

Answer:

0.0192 mL per min.

Explanation:

IV rate = 36 mg per 30 min.

IV concentration = 125 mg per 2.0 mL

36 mg per 30 min. IV rate = 36/30 = 1.2 mg per min

If 125 mg methylprednisolone is present in 2.0 mL of the IV nag, how many mL would contain 1.2 mg?

= 2x1.2/125

= 0.0192 mL

Therefore, the flow rate of the IV must be 0.0192 mL per min. in order to be able to deliver 36 mg per 30 min.

7 0

2 years ago

Compute the following and give your answer in scientific notation

lana66690 [7]

Answer:

5.7 × 10⁻³

Explanation:

3.51 - 2.94 = 0.57

Don't change the scientific notation.

0.57 × 10⁻²

BUT

The "0" is not applicable so move the decimal point to the right by one and subtract the power by one.

5.7 × 10⁻²⁻¹ = 5.7 × 10⁻³

4 0

2 years ago

Arrange the balls in order from greatest amount of gravitational potential energy to least.

Ulleksa [173]

A, b, c, f, e, d

Remember that gravitational potential energy is greatest at the top.

3 0

3 years ago

Read 2 more answers

Write balanced chemical equations for the complete neutralization reactions that take place when the following acids are titrate

swat32

Answer:

(a) $NaOH_{(aq)} + HCl_{(aq)} --> NaCl_{(aq)} + H_2O_{(l)}$

(b) $NaOH_{(aq)} + CH_3COOH_{(aq)} --> CH_3COONa_{(aq)} + H_2O_{(l)}$

Explanation:

For a reaction involving sodium hydroxide and hydrochloric acid, the balanced equation of reaction is:

$NaOH_{(aq)} + HCl_{(aq)} --> NaCl_{(aq)} + H_2O_{(l)}$

For a reaction involving sodium hydroxide and acetic acid, the balanced equation of reaction is:

$NaOH_{(aq)} + CH_3COOH_{(aq)} --> CH_3COONa_{(aq)} + H_2O_{(l)}$

For a reaction involving sodium hydroxide and phosphoric acid, the balanced equation of reaction is:

$H_3PO_4_{(aq)} + 3NaOH_{(aq)} --> 3H_2O{(l)} + Na_3PO_4_{(aq)}$

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3 years ago