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3 years ago

Can you answer this for me

Chemistry

2 answers:

MrRissso [65]3 years ago

7 0

Answer:

Algebra is an area in mathematics that uses variables, in the forms of letters and symbols, to act as numbers or quantities in equations and formulas. Geometry is an area in mathematics that studies points, lines, varied-dimensional objects and shapes, surfaces, and solids.

Explanation:

It is your own opinion on which is more difficult. In my opinion, Geometry is more difficult.

Zepler [3.9K]3 years ago

7 0

Answer: I believe they want you to know the variables; and Newton described what? Force, action, mass.

I think you should read the Laws, and see how the variables, those important pieces, are shown in an equation. That will be of real help to you. Get the very basics down in your head (and on paper), and it will be much easier!

Explanation:

You might be interested in

Which element has a crystalline lattice through which electrons flow freely

xxTIMURxx [149]

Answer:

Copper

Explanation:

Metals are often arranged in crystalline lattices. The lattice consists of metal ions with an overlying sea of electrons. The conductivity of a metal depends on the nature of this crystal lattice. Naturally, some metals conduct electricity better than others. The unique properties of the crystal lattice of copper allow electrons to flow through it freely especially when it is drawn into a wire.

5 0

4 years ago

A 1.547 g sample of blue copper(II) sulfate pentahydrate, ‍ , is heated carefully to drive off the water. The white crystals of

sweet [91]

Answer:

$=6.2x10^{-3}mol CuSO_4$

Explanation:

Hello,

By developing the following stoichiometric relationship, the required amount could be found as follows:

- Moles of $CuSO_4$ :

$1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O} =6.2x10^{-3}mol CuSO_4$

- Grams of $CuSO_4$

$1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{1mol CuSO_4}{1mol CuSO_4.5H_2O}*\frac{159.5g CuSO_4}{1mol CuSO_4} =0.989 g CuSO_4$

- Moles of water:

$1.547gCuSO_4.5H_2O *\frac{1molCuSO_4.5H_2O}{249.5gCuSO_4.5H_2O} *\frac{5mol H_2O}{1mol CuSO_4.5H_2O}=0.031mol H_2O$

Finally, one could see that the mass of the anhydrous compound is less than the pentahydrated compound since it is waterless.

Best regards.

8 0

3 years ago

Draw arrows for all of the attractive and repulsive forces between the negatively charged electrons and positively charged proto

Sophie [7]

Answer:

Attraction

(e-) ---> <--- (H+)

Repulsion:

<---(e-) (e-)-->

Neutral:

(e-) (Helium)

Explanation:

Accordingly to coulomb's law:

In the attraction, the hydrogen without an electron has a positive charge and needs to be fulfilled with a negative charge found in an eletron.

In the repulsion, both electrons has the same charge and repulse each other.

In the neutral case, the Helium is highly stable therefore the electron is not attracted.

4 0

3 years ago

A reaction generates hydrogen gas (H) as a product. The re-

Amiraneli [1.4K]

<u>Answer:</u> The average rate of the reaction is $7.82\times 10^{-3}M/min$

<u>Explanation:</u>

To calculate the molarity of hydrogen gas generated, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = $3.91\times 10^{-2}mol$

Volume of solution = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M$

Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.

To calculate the average rate of the reaction, we use the equation:

$\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}$

We are given:

Concentration of hydrogen generated = 0.1564 M

Time taken = 20.0 minutes

Putting values in above equation, we get:

$\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min$

Hence, the average rate of the reaction is $7.82\times 10^{-3}M/min$

6 0

3 years ago

Write the empirical formula

stira [4]

Answer:

$1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}$

Explanation:

$1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}$

7 0

3 years ago