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2 years ago

Pls Help! 100 Points!!!!!

Chemistry

2 answers:

Katen [24]2 years ago

6 0

Answer:

The answer is c,F has a smaller radius than F− because F− has an additional energy shell.

Explanation:

if correct mark brainliest

jarptica [38.1K]2 years ago

4 0

The answer is C.
Hope this helps:)

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A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is

vekshin1

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

$Q= m\times c\times \Delta T$

Q = heat absorbed = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C

$\Delta T={\text{Change in temperature}}=T_2-T_1$

$T_1=25^oC$

Putting values in above equation, we get:

$+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC$

Hence, the final temperature of the copper is 95°C.

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2 years ago

Which species is most vulnerable to extinction raccoon or mountain gorilla

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Mountain Gorilla’s are more vulnerable to extinction because they are endangered due to habitat loss, diseases, poaching, ect.

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3 years ago

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Explain the flexibility and the flow characteristics of water

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Explanation:

Flow-duration data are daily mean flow values measured over a specified time interval that have been exceeded various percentages of the specified time interval.

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3 years ago

If I mix bleach and namona and throw it on a person what will happen

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The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 genera

nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate = $3.45 g\times \frac{32.5}{100}=1.121 mol$

Moles of bicarbonate ion = $\frac{1.121 g/mol}{61 g/mol}=0.01840 mol$

$HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)$

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

$\frac{1}{1}\times 0.01840 mol=0.01840 mol$ of carbon dioxide gas

Moles of carbon dioxide gas n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

$PV=nRT$ (ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L$

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

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3 years ago