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2 years ago

how many grams of NaCl will be needed to form 600 ml of a saturated solution at 100 degrees celsius ?

1 answer:

7 0

The solubility of NaCl in water at 100 C is 40%, meaning that we can dissolve 40 g NaCl in 100 g water. Assuming that dissolving NaCl does not add any volume to the solution, 600 mL of water is approximately equal to 600 g of water. By ratio and proportion: 40 g NaCl/100 g H2O = x g NaCl/600 g H2O
x = 240 g NaCl
So 240 g of NaCl must be dissolved to form a saturated solution.

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What is the empirical formula of C6H18O3?

Aloiza [94]

Answer:

The answer to your question is C₂H₆O

Explanation:

Data

Molecular formula = C₆H₁₈O₃

Empirical formula = ?

Empirical formula is defined as the simplest ratio of the elements that form part of a molecule.

Process

To find the empirical formula find the greatest common factor of the subscripts.

6 18 3 2

3 9 3 3

1 3 1 3

The GCF is 3, so factor 3 of the molecular formula

3 ( C₂H₆O)

The result is the empirical formula C₂H₆O

7 0

2 years ago

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= $\frac{(CH3OH)}{(CO) x (H2)^{2}}$ =14.5

Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.

We have:

(CH3OH)= ?

(CO)= 0.15 M

(H2)= 0.36 M

So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

14.5= $\frac{(CH_{3}OH) }{(0.15 M) x (0.36 M) ^{2} }$

14.5 x (0.15 M) x $(0.36)^{2}$ = (CH₃OH)

0.2818 M = (CH₃OH)

6 0

2 years ago

Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce

Setler79 [48]

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol$

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

$Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol$

According to the reaction shown below:-

$MgCO_3\rightarrow MgO+CO_2$

$CaCO_3\rightarrow CaO+CO_2$

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = $\frac{x}{84.3139}\ mol$

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = $\frac{x}{84.3139}\times 40.3044 \ g$

Moles of CaO = $\frac{9.66 - x}{100.0869}\ mol$

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = $\frac{9.66 - x}{100.0869}\times 56.0774 \ g$

Given that total mass of the oxide = 4.84 g

Thus,

$\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84$

$\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84$

$-694.1618435x+45673.48749\dots =40843.38968\dots$

$x=\frac{4830.09780\dots }{694.1618435}$

$x=6.9582$

Thus, the mass of Magnesium carbonate = 6.9582 g

$\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100$

$\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%$

3 0

2 years ago

The information below describes a redox reaction.

Alchen [17]

Answer:

Explanation:

You have to mutiply the silver reaction by 3 in order to substract the electrons

3 0

2 years ago

Read 2 more answers

2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula

Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

C = 24%

H = 7%

F = 38%

P = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

C H F P

% composition

by mass 24 7 38 31

Molar mass 12 1 19 31

Number of

moles 24/12 7/1 38/19 31/31

2 7 2 1

Dividing

by the

smallest 2/1 7/1 2/1 1/1

2 7 2 1

Empirical formula C₂H₇F₂P

5 0

2 years ago