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solong [7]
3 years ago
15

If the point of the nail can be approximated as a circle with a radius 2.00×10^-3m What is the pressure in MPa exerted on the wa

ll if a hammer strikes the nail with a force of 104 N
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
3 0

Answer:

8.28 MPa

Explanation:

From the question given above, the following data were obtained:

Radius (r) = 2×10¯³ m

Force applied (F) = 104 N

Pressure (P) =?

Next, we shall determine the area of the nail (i.e circle). This can be obtained as follow:

Radius (r) = 2×10¯³ m

Area (A) of circle =?

Pi (π) = 3.14

A = πr²

A = 3.14 × (2×10¯³)²

A = 3.14 × 4×10¯⁶

A = 1.256×10¯⁵ m²

Next, we shall determine the pressure. This can be obtained as follow:

Force applied (F) = 104 N

Area (A) = 1.256×10¯⁵ m²

Pressure (P) =?

P = F / A

P = 104 / 1.256×10¯⁵

P = 8280254.78 Nm¯²

Finally, we shall convert 8280254.78 Nm¯² to MPa. This can be obtained as follow:

1 Nm¯² = 1×10¯⁶ MPa

Therefore,

8280254.78 Nm¯² = 8280254.78 Nm¯² × 1×10¯⁶ MPa / 1 Nm¯²

8280254.78 Nm¯² = 8.28 MPa

Thus, the pressure exerted on the wall is 8.28 MPa

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Answer:

Because it contains only two elements.

Explanation:

A binary compound is a compound made up of only two elements. Barium chloride contains:

1. Barium.

2. Chlorine.

Since Barium chloride contains only two elements, it is therefore a binary compound.

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3 years ago
Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th
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Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol

The chemical equation for the reaction of ammonia and HCl follows:

NH_3+HCl\rightarrow NH_4^++Cl^-

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonia

  • Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonium ion

  • Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.8\times 10^{-5}

10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}

The chemical equation for the dissociation of ammonium ion follows:

NH_4^+\rightarrow NH_3+H^+

The expression of K_a for above equation follows:

K_a=\frac{[NH_3][H^+]}{[NH_4^+]}

We know that:

[NH_3]=[H^+]=x

[NH_4^+]=0.072M

Putting values in above expression, we get:

5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M

To calculate the pH concentration, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=6.32\times 10^{--6}M

pH=-\log (6.32\times 10^{-6})\\\\pH=5.20

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

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The part in the middle with a coil of wire wrapped around a metal shaft is part of an electric motor. What is it?
evablogger [386]

Answer:

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Explanation:

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One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
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<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

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4 years ago
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