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3 years ago

I need help for the Newton’s of second law

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

7 0

The greater the mass the greater the force. Acceleration is produced when a force acts on a mass. If you need more, just let me know

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Murcury is the only metal at room temperature. Its density is 13.6g/mL. How many grams of murcury will occupy a volume of 95.8mL

Hatshy [7]

Answer:

<h3>The answer is 1.30288 × 10³ g</h3>

Explanation:

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

<h3>mass = Density × volume</h3>

From the question

volume = 95.8mL

density = 13.6g/mL

We have

mass = 13.6 × 95.8 = 1302.88

We have the final answer as

Hope this helps you

3 0

3 years ago

Consider the balanced equation. 2HCl + Mg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theor

Serga [27]

Answer:

Theoretical yield of hydrogen is 1.11 g

Explanation:

Balanced equation, $Mg+2HCl\rightarrow MgCl_{2}+H_{2}$

As Mg remain present in excess therefore HCl is the limiting reagent.

According to balanced equation, 2 moles of HCl produce 1 mol of $H_{2}$ .

Molar mass of HCl = 36.46 g/mol

So, 40.0 g of HCl = $\frac{40.0}{36.46}moles$ of HCl = 1.10 moles of HCl

Hence, theoretically, number of moles of $H_{2}$ are produced from 1.10 moles of HCl = $(\frac{1}{2}\times 1.10)moles=0.550moles$

Molar mass of $H_{2}$ = 2.016 g/mol

So, theoretical yield of $H_{2}$ = $(0.550\times 2.016)g=1.11g$

7 0

3 years ago

Read 2 more answers

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0

2 years ago

The downward force acting on an object in free fall is the force of

Eduardwww [97]

Gravittyyyyyyyyy ..........................
i was reported but the answer is gravity you can only be reporting if you offering 1000 points but if not please exit stage right

6 0

3 years ago

Read 2 more answers

In which of these diatomic molecules would you NOT find an octet of electrons

Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³ → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so <u>each atom has an octet of electrons.</u>

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹ → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, <em><u>we can not find an octet.</u></em>

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴ → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, <u>each oxygen atom has an octet</u>.

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵ → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, <u>each fluorine atom has an octet of electrons around</u>.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!

8 0

2 years ago