How many atoms are in a 58.2 L sample of Agent Orange gas?

ira [324]

Materials and their properties: compounds like sodium chloride - an interactive educational resource for 11 to 14 year olds. ... Elements are substances (like hydrogen and oxygen) that can't be split into simpler substances. ... For each statement, decide whether it describes a mixture or a compound and check the box.

8 0

3 years ago

How many grams of CuF2 are needed to make a 2.8 M solution

11111nata11111 [884]

Molarity is given as,

Molarity = Moles / Volume of Solution ----- (1)

Also, Moles is given as,

Moles = Mass / M.mass

Substituting value of moles in eq. 1,

Molarity = Mass / M.mass × Volume

Solving for Mass,

Mass = Molarity × M.mass × Volume ---- (2)

Data Given;

Molarity = 2.8 mol.L⁻¹

M.mass = 101.5 g.mol⁻¹

Volume = 1 L (I have assumed it because it is not given)

Putting values in eq. 2,

Mass = 2.8 mol.L⁻¹ × 101.5 g.mol⁻¹ × 1 L

Mass = 284.2 g of CuF₂

5 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

When are zeros significant in a value ?

ASHA 777 [7]

Answer:

If the zero is between fwo nonzeros

Explanation:

5 0

3 years ago

Read 2 more answers

Why does potassium explode when exposed to water?

soldi70 [24.7K]

Answer:

The highly unstable pure sodium or potassium wants to lose an electron and this splits the water atom, producing a negatively charged hydroxide ion and hydrogen and forming an explosive gas that ignites.

Explanation:

4 0

3 years ago