If you filter a mixture of sand and salt-water, what will the solid residue in the filter paper consist of?

Which the following countries founded their first permanent settlement in North America: England, the Netherlands,France,and Spa

butalik [34]

England with the Jamestown colony

8 0

3 years ago

How will you describe the behavior of dry ice?

Ganezh [65]

Answer:

Explanation: A square of dry ice has a surface temperature of - 109.3 degrees Fahrenheit (- 78.5 degrees C). Dry ice additionally has the extremely decent component of sublimation - as it separates, it transforms legitimately into carbon dioxide gas as opposed to a fluid.

6 0

3 years ago

Help me please thank you

ankoles [38]

Answer:

When nitric acid combine with sodium hydroxide the salt formed is called sodium nitrate. option B

Explanation:

It is the strong acid strong base reaction. When acid and base react with each other salt and water are formed.

In given reaction nitric acid combine with sodium hydroxide base and form sodium nitrate salt and water.

Chemical equation:

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)

Ionic equation:

H⁺NO₃⁻(aq) + Na⁺OH⁻(aq) → Na⁺NO₃⁻(aq) + H₂O(l)

Net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The Na⁺(aq) and NO₃⁻(aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3 0

2 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

6 0

2 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

2 years ago