All categories

2 years ago

Help please really need help

Chemistry

1 answer:

zheka24 [161]2 years ago

5 0

Answer:

2.14 × 10⁻³ molecules/RSP

3.31 × 10⁻³ molecules/ESP

Explanation:

Step 1: Calculate the number of moles of Acetaminophen per Regular Strength Pill (RSP)

A Regular Strength Pill has 1.29 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.29 × 10²¹ molecules/RSP × 1 mol/6.02 × 10²³ molecules = 2.14 × 10⁻³ molecules/RSP

Step 2: Calculate the number of moles of Acetaminophen per Extra Strength Pill (ESP)

An Extra Strength Pill has 1.99 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.99 × 10²¹ molecules/ESP × 1 mol/6.02 × 10²³ molecules = 3.31 × 10⁻³ molecules/ESP

You might be interested in

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

Kamila [148]

The answer for the following problem is mentioned below.

Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

Final pressure of the gas ( $P_{2}$ ) = 4 atm

Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

So;

we can tell from the above equation;

P ∝ T

(i.e.)

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

$T_{2}$ = 740 K

Therefore the final temperature of the gas is 740 K

8 0

3 years ago

Predict how the addition of a catalyst would affect the rate of the reaction below, and explain your prediction H2 (g) + I2 (g)

Fynjy0 [20]

A catalyst is a chemical substance that hastens the chemical reaction. This does not participates in the creating the product(s) but allows it to be formed easily. With this, it is now known that the rate of the reaction becomes relatively higher compared to the uncatalyzed reactions.

Therefore, the answer to this item is the rate of the reaction becomes faster.

4 0

2 years ago

Read 2 more answers

Write the electron configuration for a sodium cation with a charge of 1+

Dafna1 [17]

$1{s}^{2} 2 {s}^{2} 2 {p}^{6}$

or simply 2,8 it is isoelectronic with argon

4 0

3 years ago

An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i

lutik1710 [3]

Answer:

$C_{metal}=126.6\frac{J}{g\°C}$

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

$Q_{metal}=-Q_{water}$

Which can be also written as:

$m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})$

Thus, since we need the specific heat of the metal, we solve for it as shown below:

$C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}$

Best regards.

7 0

2 years ago

WILL MARK BRAINLY

DENIUS [597]

Answer:

its B i guess

Explanation:

7 0

2 years ago