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andreev551 [17]

3 years ago

7

**Suppose a cart is being pushed by a certain net force. If the net force is doubled, by how much does the acceleration change?

Explain.
NO WORK = NO CREDIT = REPORT

thx <3

1 answer:

White raven [17]3 years ago

4 0

Yes because the force or the pressure of the cart is getting doubled therefore it makes the speed or the net force harder

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Match the layers of the atmosphere

Ira Lisetskai [31]

Answer:

What language is this please i need help

7 0

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

A solution of salt water is boiled. The water evaporates and solid salt remains. Which statement is true?

Scorpion4ik [409]

C because neither the water or salt have changed their molecular structure. H2O is still H2O and NaCl is still NaCl.

4 0

4 years ago

Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas , ch

FinnZ [79.3K]

Answer:

2.9 g of water are produced by 9.6 g of ammonium perchlorate

<em>Note: The question is incomplete. The complete question is given below:</em>

<em>Ammonium perchlorate (NH₄ClO₄) is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) , and a great deal of energy. What mass of water is produced by the reaction of 9.6 g of ammonium perchlorate? Be sure your answer has the correct number of significant digits.</em>

Explanation:

The reaction of ammonium perchlorate (NH₄ClO₄) to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) is shown in the balanced chemical equation given below:

2 NH₄ClO₄ → N₂ + Cl₂ + 2 O₂ + 4 H₂O

From the equation, 2 moles of ammonium perchlorate produces 4 moles of water, i.e. mole ratio of NH₄ClO₄ to H₂O = 2 : 4 = 1 : 2

molar mass of ammonium perchlorate, NH₄ClO₄ = (14 + 4 * 1 + 35.5 +16 * 4) = 117.5

molar mass of water, H₂O = (2 * 1 + 16) = 18.0 g

mass of water produced = moles of ammonium perchlorate * 2 * molar mass of water

moles of perchlorate = mass / molar mass = 9.6/117.5

mass of water produced = 9.6/117.5 * 2 * 18.0 g = 2.94 g of water

Therefore, 2.9 g of water are produced by 9.6 g of ammonium perchlorate

6 0

3 years ago

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decomposes after 5.55 days if the sample i

Leni [432]

Answer:

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/ $2^{5.55/3.823}$

R.R = 100/ $2^{1.45}$

R.R = 100/2.73

R.R = 36.63 g.

The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

= 63.37 ≈ 63.4 g

The mass of radon that decompose = 63. 4 g

8 0

3 years ago