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3 years ago

What does each of the facts shown above tell you about the element on the periodic table

1 answer:

7 0

<span>The atomic number of a chemical element (also known as its proton number) is the number of protons found in the nucleus of an atom of that element, It can be used to determining which group the element belongs to in the periodic table. Therefore, atomic number is the correct answer.</span>

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Aluminum has a density of 2.7g/cm how much space in cm 3 would 100 grams of aluminum occupy?

PIT_PIT [208]

Answer:

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Explanation:

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3 years ago

What is the charge and location of a neutron?

Whitepunk [10]

Answer:

There is no charge and it is in nucleus

Explanation:

the neutron is the nuetral one and the protron is positive charger and electron is negative. please let me know if you need me to explain more

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4 years ago

Read 2 more answers

10. Define experiment.

Ber [7]

Answer:

to make a discovery, test a hypothesis, or demonstrate a known fact.

Explanation:

perform a scientific procedure, especially in a laboratory, to determine something.

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3 years ago

Read 2 more answers

What types of R groups can exist in amino acids? (Choose all that apply)

PilotLPTM [1.2K]

Answer:

acidic,non polar,basic

Explanation:

that’s the ones I put and I got a 100% on my test just now

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3 years ago

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

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3 years ago