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3 years ago

Some forms of vitamin D, C₂₈H₄₄O, can be found in red meat. How many total atoms are in 5 molecules of C₂₈H₄₄O

Chemistry

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

3.01 × 10^24 atoms of vitamin D

Explanation:

The number of atoms, molecules or ions present in a substance is given by the Avogadro's number which is 6.02 × 10^23.

Hence;

1 molecule of vitamin D contains 6.02 ×10^23 atoms

5 molecules of vitamin D contains 5 × 6.02 ×10^23/1

= 3.01 × 10^24 atoms of vitamin D

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HURRY PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!

aivan3 [116]

Answer:

When a scientist on Earth drops a hammer and a feather at the same time an astronaut on the moon drops a hammer and a feather, the result expected is that the hammer hits the ground before the feather on Earth, and the hammer and feather hit at the same time on the moon (option D).

Explanation:

In the abscence of atmosphere (vacuum), the objects fall in free fall. This is, the only force acting on the objects is the gravitational pull, which is directed vertlcally downward.

Under such absecence of air, the equations that rules the motion are:

V = Vo + gt
d = Vo + gt² / 2
Vf² = Vo² + 2gd

As you see, all those equations are independent of the mass and shape of the object. This explains why when an astronaut on the moon drops a hammer and a feather at the same time, the hammer and feather hit at the same time on the moon, a space body where the gravitational attraction is so small (approximately 1/6 of the gravitational acceleration on Earth) that does not retain atmosphere.

On the other hand, the air (atmosphere) present in Earth will exert a considerable drag force on the feather (given its shape and small mass), slowing it down, whereas, the effect of the air on the hammer is almost neglectable. In general and as an approximation, the motion of the heavy bodies that fall near the surface is ruled by the free fall equations shown above, so, the result that is expected when a scientist on Earth drops a hammer and a feather at the same time is that the hammer hits the ground before the feather.

8 0

3 years ago

Read 2 more answers

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what

ludmilkaskok [199]

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

m = mass
c = specific heat capacity
∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

8 0

2 years ago

Can anyone teach me what is criss cross method in chemistry

Solnce55 [7]

Answer:

In the criss-cross method, the numerical value of the ion charge of the two atoms are crossed over, which becomes the subscript of the other ion. Using this technique, we will write the chemical formula of the given compounds.

Criss cross the absolute values to give Al2O3. To find the formula for magnesium oxide:- The oxidation number of Mg is +2 and oxygen is -2. Criss cross the absolute values to give Mg2O2In this example there is a common factor of 2 so divide by 2 to give MgO.

8 0

2 years ago

Determine the number of moles of gas in a 4.1L container at 341K with a pressure of 1.78 atm.

lozanna [386]

Answer:

0.26 mol

Explanation:

using general gas equation

PV=nRT

V=4.1litre= 4.1 dm³

P= 1.78 atm

R= 0.0821

PUT VALUES

8 0

3 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago