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1 year ago

Identify the charge on the atom or ion that has one proton, zero neutrons, and zero electrons.

Chemistry

1 answer:

MAVERICK [17]1 year ago

3 0

Answer:

The emission spectrum of gold shows a line of wavelength 2.676 x 10-7 m. How much energy is emitted as the excited electron falls to the lower energy level?

(a) 7.43 x 10-19 J

(b) 5.30 x 10-20 J

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Define saturated and unsaturated fats

saul85 [17]

Answer:

SATURATED: Saturated fat is a type of fat in which the fatty acid chains have all or predominantly single bonds.

UNSATURATED: Unsaturated fats are loosely packed. They tend to be liquid at room temperature.

There are two main types of unsaturated fat:

Monounsaturated

Polyunsaturated

8 0

2 years ago

According to the graph, which statement is true?

damaskus [11]

Nitrogen is the one

5 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

An aqueous solution is found to be 3.64% potassium nitrate by mass, how many grams of solution should I evaporate the water from

Sidana [21]

Answer:

513.74 g of solution

Explanation:

% Mass grams are defined as the grams that are dissolved in salt (in this case, it would be potassium nitrate) dissolved every 100 g of the solution. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.

The relationship is:

3.64 g of potassium nitrate _____ 100 g solution

18.7 g of potassium nitrate _____ X = 513.74 g of solution

Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution

So, I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it.

8 0

2 years ago

Write a word equation for the extraction of tin from tin oxide using carbon

Ann [662]

SnO2 + 2C => Sn + 2CO

3 0

2 years ago