Ask question

Ask question

All categories

4 years ago

5

A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution? Us

e M subscript i V subscript i equals M subscript f V subscript f..

1 answer:

drek231 [11]4 years ago

3 0

Answer:6.0M

Explanation:see attached photo

You might be interested in

Magnesium has three naturally occurring isotopes with masses of 23.99 amu, 24.99 amu, and 25.98 amu and natural abundances of 78

mihalych1998 [28]

Answer:24.31

Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97

Contribution made by isotope of mass 24.99=24.99×10.00=249.9

Contribution made by isotope of mass 25.98=25.98×11.01=286.04

Total contribution=1894.97+249.9+286.04=2430.91

Average mass=2430.91÷100

=24.31

3 0

3 years ago

What happens when energy is removed from liquid water

mariarad [96]

Answer:

When energy is removed in liquid water then it will solidify since heat is given off by the system to its surrounding. It is an exothermic process where the enthalpy decreases since heat is taken off. So liquid water becomes ice in an exothermic process.

Explanation:

8 0

3 years ago

To find the atomic mass of a particular atom you need to know

Andreyy89

You just need the number of protons and number of neutrons as the mass of eelctrons is negligible

5 0

3 years ago

Read 2 more answers

Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of chromium(III) bromide and pota

expeople1 [14]

Explanation:

An ionic equation will be the one in which all the participating species will be present as ions.

The given reaction will be as follows.

$CrBr_{3} + K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + KBr$

Balancing this equation by multiplying $CrBr_{3}$ by 2 and $K_{2}CO_{3}$ by 3 on reactant side. Whereas multiply KBr by 6 on product side.

$2CrBr_{3} + 3K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6KBr$

Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) + 6K^{+} + 6Br^{-} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6K^{+} + 6Br^{-}$

As both $K^{+}$ and $Br^{-}$ are spectator ions. Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) \rightarrow Cr_{2}(CO_{3})_{3}(s)$

4 0

3 years ago

What is the electron configuration of argon

PilotLPTM [1.2K]

Atomic number - 18
1s2,2s2,sp6,3s2,3p6
noble configuration is [Ne]3s2,3p6

7 0

3 years ago