All categories

4 years ago

In the solubility lab, what did you change the temperature of

1 answer:

6 0

I'm taking the test right now; I'll edit this when I find out if I was right! (I chose "solvent" but I wanna be sure.)

Edit: Solvent is the correct answer!

You might be interested in

3. What do the most abundant elements in Earth's atmosphere have in common?

Advocard [28]

Answer:

They all have: nitrogen :780,900 755,100

oxygen: 209,500 231,500

argon 9,300 12,800

carbon dioxide

386 591

8 0

3 years ago

This number tells you the number of each type of atom in a compound.

KonstantinChe [14]

Answer:

avogadro's constant

Explanation:

this is the fixed number of the atoms in the molecule of an element

avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

that is all gases with same temperature and pressure will always have equal numbers of molecules

7 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

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

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

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

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

Which of the following isotopes is NOT considered to be a major threat to human health from nuclear fallout? a. strontium-90 b.

Nat2105 [25]

Ohhh it’s A) strontium

4 0

3 years ago

What’s the answer to the question on the right

just olya [345]

All three of them i know i already did that

7 0

3 years ago