Answer : 5.405 moles of
are equivalent to 97.3 grams of 
Solution : Given,
Mass of
= 97.3 g
Molar mass of
= 18 g/mole
Formula used :

or,

Now put all the given values in this formula, we get the moles of 

Therefore, 5.405 moles of
are equivalent to 97.3 grams of 
<h2>
Answer:C</h2>
Explanation:
A chemical equation is a equation that describes a corresponding chemical reaction.
A chemical reaction is generally written as
→
refer to all the reactants involved in the chemical reaction.
Reactants are usually written on the left hand side of the chemical equation.
refer to all the products formed in the chemical reaction.
products are usually written on the right hand side of the chemical equation.
In the given reaction,
,
are written on the right side of the equation.
So,
,
are the products.
Answer:
Beta decay is most common in elements with a high neutron to proton ratio. Gamma decay follows the form: In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state.
Explanation:
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Answer:
Here's what I get
Explanation:
1. Sugar
(a) Dissolving in water
The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.
2. Magnesium sulfate
(a) Dissolving in water
The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.
You can't usually just use a single spectrum line to confirm the identity of an element because there are cases that the emission line id not clearly defined. When the emission line is very weak compared to surrounding noise, in which case the more datapoints you have to build up confidence for the existence of a particular emission spectra, the better.