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Luda [366]
3 years ago
15

Jamie completes the following steps in an experiment to investigate how temperature affects how fast something will rust in wate

r. He places a nail in a bowl of water in the sun and a nail in a bowl of water in the shade. He also leaves an iron nail out of the water in a dry bag out of the sun. Why would Jamie leave one out of the water and out of the sun?
A Jamie also wants to test if the water affects how fast the iron nail will rust.
B Jamie notices that the size of the nail is different, so he does not want to use it later.
C Jamie doing all these steps to create a control group to compare his results to.
D Jamie puts the nail in a cooler place to see if it will rust as fast as the other items.
Chemistry
2 answers:
stiks02 [169]3 years ago
6 0
B. Jamie notices that the size of the nail is different, so he does not want to use it later
earnstyle [38]3 years ago
5 0

Answer:

I think B

Explanation:

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How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi
Cerrena [4.2K]

Answer:

6.7511\times 10^7\ C

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains 6.02\times 10^{23} atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains \frac {6.02\times 10^{23}}{15.9994}\times 1400 atoms of oxygen.

Number of atoms in 1400 g of oxygen = 526.769754\times 10^{23}

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + 1.602\times 10^{-19}\ C

So, Charge on 1 atom of oxygen = 8\times 1.602\times 10^{-19}\ C

Thus,

Charge on 526.769754\times 10^{23} atoms of oxygen = 526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C

Thus, positive charge in 1.4 kg of oxygen = 6.7511\times 10^7\ C

5 0
3 years ago
Please help I’ll give brainliest
jonny [76]

REACTION TYPE:

Single displacement/Oxidation-reduction

(Hope this helped UwU)

4 0
3 years ago
Read 2 more answers
A 410 L volume of nitrogen gas is cooled from 61 C to -25 C. What is the volume of the nitrogen at the lower temperature if all
Lena [83]

Answer:

304.5\text{L}

Explanation:

Here, we want to get the volume of the nitrogen gas at the lower temperature

From Charles' law, we know that volume and temperature (in Kelvin) are directly proportion

The mathematical relationship is:

\frac{V_1}{T_1}\text{ = }\frac{V_2}{T_2}

Where:

V1 is the initial volume which is 410 L

V2 is the final volume which is unknown

T1 is the initial temperature which we will convert to Kelvin by adding 273.15 K, we have it as 61 + 273.15 = 334.15 K

T2 is the final temperature which we have to convert to Kelvin by adding it to 273.15K : We have that as -25 + 273.15 = 248.15 K

Substituting the values, we have it that:

\begin{gathered} \frac{410}{334.15}\text{ = }\frac{V_2}{248.15} \\  \\ V_2\text{ = }\frac{248.15\text{ }\times410}{334.15} \\  \\ V_2\text{ = 304.5 L} \end{gathered}

6 0
1 year ago
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f
deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

           S = K_{b} \times ln \Omega

where,      S = entropy

             K_{b} = Boltzmann constant

             \Omega = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that \Omega = 1, and \Omega' = 0.833

Therefore, change in entropy will be calculated as follows.

     \Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega

                 = 1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)

                 = 1.38 \times 10^{-23} \times (-0.182)

                 = -0.251 \times 10^{-23}

or,             = -2.51 \times 10^{-24}

Thus, we can conclude that the entropy change for a particle in the given system is -2.51 \times 10^{-24} J/K particle.

8 0
3 years ago
Part 1: Draw the major organic product for the proton transfer reaction between sodium hydride and ethanol; be sure to include l
valina [46]

Answer:

Major product ethoxide ion

Explanation:

  • Sodium hydride acts as a strong base towards ethanol.
  • Hydride ion abstracts one proton from -OH group in ethanol to produce sodium ethoxide and hydrogen gas.
  • It is an example of acid-base reaction where sodium hydride acts as a base and ethanol acts as an acid
  • Structure of major organic product i.e. ethoxide ion has been shown below.

4 0
3 years ago
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