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3 years ago

Using the Erlenmeyer flask and its contents from part A, continue with

Chemistry

2 answers:

dangina [55]3 years ago

8 0

Answer:

negative is it can't be broken down and makes the air polluted. Positive, it can be re used again and again.

Explanation: here is you answer please rate me the most brainlest

Anestetic [448]3 years ago

5 0

It’s b for sure that’s the answer

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If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?

NARA [144]

I have attached a photo of the structure.
You can get better at solving problems like this by practicing a lot!

5 0

3 years ago

A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w

Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

$Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)$

$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

$\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol$

4 0

3 years ago

i have a crossword thing that i got from school and one of the clues is: If there are two equal and opposite forces on something

cluponka [151]

I would assume the answer is called attractive

6 0

3 years ago

Read 2 more answers

Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

lina2011 [118]

Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

Explanation: We are given $k_b$ of $C_6H_5NH_2$ at 25°C which is $1.7\times 10^{-9}$

To calculate the $k_a$ of $C_6H_5NH_3^+$ , we use the formula:

$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

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3 years ago

How much NaNO3 can I dissolve in 400 grams of water at 283 K?

jeyben [28]

929282827280182292848474748393027373882

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2 years ago