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pashok25 [27]
3 years ago
10

What is 4 2/3 - 1 3/5=

Chemistry
1 answer:
rjkz [21]3 years ago
3 0

4 2/3-1 3/5= 46/15. Which is 3.06666666666 (Continuing) as a decimal.

3 1/15 as a mixed number

-TheOneandOnly.

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My science teacher trying to be cool
Helga [31]

Answer:

omg lol XD

Explanation:

3 0
2 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Draw the Lewis structures of the molecules below and use them to answer the following questions:
Gnesinka [82]

Answer:

If NO₂ molecule written is for Nitrogen dioxide, then, four of the five molecules presented above have no dipole moment and only one of the five molecules, Ozone (0₃), has a dipole moment.

But if the NO₂ molecule is for nitrite ion, NO₂⁻, then three out of the five molecules presented have no dipole moment and only the Nitrite ion, NO₂⁻, and Ozone, 0₃, have dipole moments.

Explanation:

- The Lewis Structure for the molecules are drawn in the image attached to this answer.

The bond dipole moment uses the idea of electric dipole moment to measure the polarity of a chemical bond within a molecule. It occurs whenever there is a separation of positive and negative charges. Polarity occurs due to differences in electronegativity.

1) Browne or Trihydridoboron, BH₃ - No dipole moment in the molecule.

Each B-H bond in BH₃ is polar/forms a dipole because the B and H atoms have different electronegativities. But, the shape of the molecule is trigonal planar which is symmetrical, so the dipoles/bond polarities cancel. The resulting BH₃ molecule is non-polar.

2)Nitrogen dioxide, NO₂ has no dipole moment.

Nitrite ion, NO₂⁻ -> Has a dipole moment.

There are two NO₂ molecules, the Nitrogen dioxide molecule is linear and has no dipole moment, but the NO₂⁻ ion is a polar molecule. The geometry of the molecule is bent because of a non-bonding pair of electrons. The bent geometry causes the polarity and subsequent dipole moment.

3) Sulfur hexafluoride, SF₆ - no dipole moment.

Sulfur hexafluoride, abbreviated as SF₆, is a nonpolar molecule. SF₆ has an octahedral molecular geometry, which means that the sulfur molecule has six fluorine atoms surrounding it. While each individual bond is polar, there is no net effect as symmetrical nature of this octahedral molecular structure means the dipole moments all cancel out, meaning that the molecule is nonpolar.

4) Ozone, O₃ - has a dipole moment.

O₃ is polar because there are 18 valence electrons, so the lewis structure would position the central O connected to one single bond and one double bond to connect the other O's. The lone pair on the central O would also mean the molecule was bent, thus making it polar. Therefore, Ozone is a polar molecule with a dipole moment of 0.53 D. The molecule can be represented as a resonance hybrid with two contributing structures, each with a single bond on one side and double bond on the other.

5) Phosphorus pentachloride, PCl₅

PCl₅ has a symmetrical geometry, the vector sum of the dipole moments of the different P-Cl bonds cancel each other. Therefore, the overall dipole moment of PCl₅ becomes 0.

Hope this helps!

6 0
3 years ago
The difference in an area with high concentration and an area with low concentration is called.
Varvara68 [4.7K]

The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.

This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

Learn more about diffusion here ;

brainly.com/question/24746577

#SPJ1

5 0
2 years ago
How do anion formation and valence electrons relate?
vagabundo [1.1K]

Answer:

A cation is formed when a metal ion loses a valence electron while an anion is formed when a non-metal gains a valence electron. They both achieve a more stable electronic configuration through this exchange.

7 0
3 years ago
Read 2 more answers
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