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Ymorist [56]
3 years ago
13

What elements are found in salt?

Chemistry
1 answer:
devlian [24]3 years ago
8 0
Sodium is a member of the alkali metal family with potassium (K) and Lithium (LI) sodium's big claim to fame is that it's one or two elements in your table salt. when bonded to chlorine (CI) THE two elements make sodium chloride
You might be interested in
A 25.0 g block of a metal alloy at 100 C is dropped into an insulated flask containing 110 g of ice; 5.30 g of the ice melted. W
mr_godi [17]

Answer:

0.00353J/g/°C

Explanation:

I will assume the temperature of the ice to be approximately 0°C.

Moreover, Heat of fusion of water is 6kJ

Amount of heat used to melt 5.3g of ice = 5.3 x 6 / 18

=1.767g°C

Therefore

1.767 = 25 x specific heat cap. x 200

Specific heat cap. = 1.767/(25x200)

= 0.00353J/g/°C

6 0
3 years ago
At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

8 0
3 years ago
Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox
Nina [5.8K]

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

3 0
3 years ago
How many moles of H2 can be made from complete reaction of 3.0 moles of Al?
tresset_1 [31]

Answer: 4.5 moles of H_{2} can be made from complete reaction of 3.0 moles of Al.

Explanation:

The given reaction equation is as follows.

2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}

This shows that 2 moles of Al reacts with 6 moles of HCl. So, the amount of HCl required to react with 1 mole Al is three times the amount of HCl.

Therefore, 3 moles of Al will react with 9 moles of HCl to give 3 moles of AlCl_{3} and \frac{9}{2} moles of H_{2}.

The reaction equation now will be as follows.

3Al + 9HCl \rightarrow 3AlCl_{3} + \frac{9}{2}H_{2}

The moles \frac{9}{2} can also be written as 4.5 moles.

Thus, we can conclude that 4.5 moles of H_{2} can be made from complete reaction of 3.0 moles of Al.

6 0
3 years ago
A chemistry student’s height is measured at 68.5 inches. How tall is the student in cm?
pantera1 [17]
173.99cm multiply 68.5 x 2.54 always use that number when converting inches to cm
4 0
3 years ago
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