Ask question

Ask question

All categories

3 years ago

11

Organic molecules always:

1 answer:

trapecia [35]3 years ago

8 0

Organic chemistry is the study of molecules containing carbon, so A

You might be interested in

fish give off the compound ammonia which has a pH above 7 to which class of compounds does ammonia belong

jasenka [17]

<span><u><em>Answer:</em></u>
The compound ammonia given by fish is alkaline

<u><em>Explanation:</em></u>
<u>We can classify elements/compounds based on their pH values into three types:</u>
<u>acids:</u> these are compounds having pH value lower than 7
<u>neutral:</u> these are compounds having pH value equal to 7
<u>alkalies:</u> these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)</span>

5 0

2 years ago

How many molecules are there in 345 grams of carbon dioxide?

Mekhanik [1.2K]

You take the grams of CO₂ times Avogadro's number divided by the molar mass.

6 0

2 years ago

Read 2 more answers

Please help me please

sweet-ann [11.9K]

Explanation:

1 sugar solution =Distillation

2 Iron powder and sand=magnetic separation

3 petrol and diesel= Fractional distillation

4 Camphor and glass powder = Sublimation

7 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago

matter is passed down from the living part of an ecosystem to the non living part when witch of the following take place ?

Ivenika [448]

I don’t know about that

7 0

2 years ago