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liraira [26]
3 years ago
11

10g of an unknown compound are added to water to form a 7.89 molar solution. if 2 liters of solution are present, what is the mo

lar mass of the unknown compound?
Chemistry
1 answer:
Citrus2011 [14]3 years ago
6 0
10 / (7.89) (2) 

molarity equation is M = Moles of solute / Liters of solution

7.89 = Moles of solute / 2

(7.89)(2). The question asks for the molar mass which is defined as g/mol. Dividing would give you the unknown compound. 

<span>MM = 10g / (7.89)(2)</span>
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Answer:- 2.39 mL are required.

Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:

M_1V_1=M_2V_2

Where, M_1 and M_2 are molarities of concentrated and diluted solutions and V_1 and V_2 are their respective volumes.

M_1 = 1.10M

M_2 = 5.00mM = 0.005M    (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)

V_1 = ?

V_2 = 525 mL

Let's plug in the given values in the formula:

1.10M(V_1)=0.005M(525mL)

V_1=(\frac{0.005M*525mL}{1.10M})

V_1=2.39mL

So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.

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3 years ago
Mario places 10 mL of water in a test tube and heats the liquid over a Bunsen burner for 2 minutes. After removing the test tube
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Mario places 10 mL of water in a test tube and heats the liquid over a Bunsen burner for 2 minutes. After removing the test tube from the Bunsen burner, there are 6 mL of water left in the test tube. This experiment is a good example of a <span>physical change involving phase changes. </span>
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3 years ago
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Which statement is true about the electrons that can be located together in an orbital?
Salsk061 [2.6K]

The two electrons that share an orbital repel each other.

All electrons bear a negative charge. They are held in their orbits by the attractive force of charged protons. The farther away an orbital is to the atomic nucleus the easier it is to expunge an electron from this distant orbital shell.

Explanation:

Because electrons have the same negative charge, they repel each other especially when they occupy the same orbital shell in an atom. To reduce this repulsion, each of the electrons in the orbital shell (remember electrons occupy orbital shells of atoms in 2s) assumes an opposite quantum (M<em>s</em>) spin; one with  – ½ while the other + ½ .

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brainly.com/question/13251728

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3 years ago
Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
Fantom [35]

Answer:

Phosphorus-28 undergoes beta-minus decay to produce

  • an electron,
  • a Silicon-28 nuclei, and
  • an electron antineutrino.

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

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The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

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<u>Answer:</u>

I think it's (C)

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