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3 years ago

When 2.50grams of nacl is dissolved in 2liters of water what is the concentration of the solutuon

Chemistry

1 answer:

blondinia [14]3 years ago

8 0

The concentration of the solution is also known as Molarity (M). The formula for molarity is

M = mol / liters

We already have the liters, so we only need to find moles. We can find the number of moles by using the molar mass of NaCl and the given amount. That can be found using the elements' atomic weights given on the periodic table.

22.990 (mass of Na) + 35.453 (mass of Chlorine) = 58.443 g / mol

That's how many grams are in one mole of NaCl, but we only have 2.5 grams. So, let's divide and see how many moles we have.

2.5 / 58.443 = .042776 moles

Now we can plug our numbers into the formula.

$M = \frac{mol}{liters}$

$M = \frac{.042776 mols}{2 liters}$

M = .02138 mols/L

So, your concentration is about .02 mol/L.

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<u>Answer:</u>

<u>For A:</u> The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

<u>For B:</u> The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

<u>For C:</u> The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

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3 years ago

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3 years ago

Chem quiz please help

GalinKa [24]

The theoretical and percentage yield for the reaction are:

The theoretical yield is 21 g
The percentage yield is 119%

<h3>Balanced equation </h3>

CH₄ + 2O₂ —> CO₂ + 2H₂O

Molar mass of CH₄ = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 44 g of CO₂

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂

Therefore,

20 g of CH₄ will react with = (20 × 64 ) / 16 = 80 g of O₂

From the above calculation, a higher mass (i.e 80 g) of O₂ than what was given (i.e 30 g) is needed to react completely with 20 g of CH₄.

Therefore, O₂ is the limiting reactant

<h3>How to determine the theoretical yield of CO₂</h3>

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂

Therefore,

30g of O₂ will react to produce = (30 × 44) / 64 = 21 g of CO₂

<h3>How to determine the percentage yield </h3>

Actual yield of CO₂ = 25 g
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Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (25 / 21) ×100

Percentage yield = 119%

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(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

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(e) S²⁻ (charge 2-)

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Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

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I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

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3 years ago

When 2.50grams of nacl is dissolved in 2liters of water what is the concentration of the solutuon​

When 2.50grams of nacl is dissolved in 2liters of water what is the concentration of the solutuon