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Semmy [17]
3 years ago
8

What is the difference between an atom and an element? a An atom cannot be cut into smaller, but an element is a group of atoms

that are the same b An atom makes up matter but an element does not c An atom can be combined but an element cannot be combined d An atom can be cut apart but an element cannot
Chemistry
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

c.

Explanation:

One of the main differences between an atom and an element is that an atom can be combined but an element cannot be combined. There are many combinations of atoms that make up different gases, liquids, and solids each with a unique makup. For example, water is made up of two hydrogen atoms and one oxygen atom (H20). Elements are made up of only the same type of atom. For example, the element Hydrogen can only contain hydrogen atoms, while the element Carbon can only contain carbon atoms.

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a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

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

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
HELP! ASAP!!! WILL MARK BRAINLIEST!!
san4es73 [151]

Answer:

Rb

Alkali Metals are Group 1 so

Rb it isnt Hydrogen because it is a gas

Explanation:

8 0
3 years ago
Read 2 more answers
If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula
DedPeter [7]

Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.

A rate law indicates the rate of a chemical response depends on reactant concentration. For a response inclusive of the price regulation commonly has the form rate = ok[A]ⁿ, in which okay is a proportionality constant known as the fee regular and n is the order.

The charge of a chemical response is, perhaps, its maximum crucial asset because it dictates whether or not a reaction can arise all throughout an entire life. knowing the charge regulation, an expression concerning the price to the concentrations of reactants can assist a chemist to modify the response conditions to get an extra suitable rate.

half-life is the time taken for the radioactivity of a substance to fall to 1/2 its authentic cost whereas implies existence is the common life of all the nuclei of a particular risky atomic species.

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3 0
2 years ago
Why does ice get colder when you melt it?
galina1969 [7]
It doesn't?

Heat transfers from hot objects to cold objects and for ice to melt it has to increase the temperature.
8 0
3 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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