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3 years ago

What are the variables of Boyle's law?

Chemistry

1 answer:

lianna [129]3 years ago

7 0

Boyles law is P1V1=P2V2 where P is pressure and V is volume so the answer is pressure and volume

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Which element has atoms that can bond with each other to form ring, chain, and network structures?

matrenka [14]

This is basics of organic chemistry - the element that has atoms that can bond with each other to form ring, chain, and network structures is (3) carbon.
The other elements available above cannot form such structures, whereas carbon, which is abundant in nature, can.

6 0

4 years ago

Read 2 more answers

5.) You bought a new car and estimated that your monthly payment would be $312. However, your actual

Dafna11 [192]

13 was the correct amount

7 0

3 years ago

The density of ice is 0.917 g/cm3. What volume of ice does 52.75 g occupy?

Vadim26 [7]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass of ice = 52.75 g

density = 0.917 g/cm³

The volume of the ice is

$volume = \frac{52.75}{0.917} \\ = 57.52453653$

We have the final answer as

Hope this helps you

4 0

3 years ago

Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of lithium sulfide and copper (II)

Aloiza [94]

Answer:

$2Li^+_{(aq)}+S^{2-}_{(aq)}+Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}\rightarrow CuS_{(s)}+2Li^+_{(aq)}+2NO_3^{-}_{(aq)}$

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

The balanced molecular equation will be,

$Li_2S_{(aq)}+Cu(NO_3)_2_{(aq)}\rightarrow CuS_{(s)}+2LiNO_3_{(aq)}$

The complete ionic equation in separated aqueous solution will be,

$2Li^+_{(aq)}+S^{2-}_{(aq)}+Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}\rightarrow CuS_{(s)}+2Li^+_{(aq)}+2NO_3^{-}_{(aq)}$

4 0

3 years ago

Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be

pashok25 [27]

Answer:

3.336.

Explanation:

<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>

<em />

So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>

<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>

∵ pH = - log[H⁺].

<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>

7 0

4 years ago