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prisoha [69]
2 years ago
14

A graph with horizontal axis time (seconds) and vertical axis position (meters). An orange line runs straight upward from 0 seco

nd zero meters. A concave blue line runs upward from the same point. Maia says that both lines on this position vs time graph show acceleration. Is she correct? Why or why not?
Chemistry
2 answers:
Pavlova-9 [17]2 years ago
7 0

Answer:

Maia is not correct because a straight line moving upwards on a position vs time graph has a constant slope, so it represents constant velocity. No change in velocity means no acceleration is taking place. Is the correct answer! :)

Explanation:

The Person/ People in the comments is correct!

OLga [1]2 years ago
6 0

Answer:

Let's start by using the definition of acceleration. Acceleration is defined as the change in velocity over the change in time. In equation, that would be Δvelocity/Δtime. Based on the axes of the given graph, it shows the trend of position over time. So, the slope of the line and the curve shows the change of position over change of time, Δdistance/Δtime. In physics, this is the definition of speed or velocity. So, Maia is incorrect. Both curves show the speed or velocity of the object, and not acceleration.  If the graph used a y-axis of velocity instead of position, then only at that instance, would be Maia be correct.

The difference between the two is, the straight line shows constant velocity while the curve line shows changing velocity.

Explanation:

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How many sodium ions are present in 325 ml of 0.850 m na2so4?
NISA [10]

Answer:

Na^+_{ions}=3.33x10^{23}Na^+_{ions}

Explanation:

Hello,

In this case, for the given molarity and volume of such solution, the moles of sodium sulfate are computed below:

n_{Na_2SO_4}=0.850\frac{mol}{L}*0.325L=0.276molNa_2SO_4

Now, by using the Avogadro's number, the ions result:

Na^+_{ions}=0.276molNa_2SO_4*\frac{2molNa^+}{1molNa_2SO_4} *\frac{6.022x10^{23}Na^+_{ions}}{1molNa^+} \\\\Na^+_{ions}=3.33x10^{23}Na^+_{ions}

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4 0
2 years ago
Identify factors that account for the high phosphoryl‑transfer potential of the nucleoside triphosphates, NTPs.
lesantik [10]

Answer:

B) stabilization by hydration

C) resonance stabilization

E) increase in entropy

Explanation:

The high phosphoryl potential of ATP results from structural differences that exist between ATP and it's product of hydrolysis. There is higher phosphoryl transfer potential from ATP than glycerol 3-phosphate.

There are some factors associated to the high phosphoryl-transfer potential of ATP which are;

1.)Electrostatic repulsion

2.) Resonance stabilization

3.) Increase in entropy.

4. Stabilization by hydration.

ATP has a phosphoryl-transfer potential which lyes between high phosphoryl-potential compounds that is a derivation of fuel molecules and acceptor molecules that needs the adequate addition of a phosphoryl group for cellular needs.

8 0
2 years ago
Which set contains no ionic species?
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The awnser should be A
7 0
2 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

        = -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

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8 0
2 years ago
At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicat
umka21 [38]

Answer:

S°m,298K = 85.184 J/Kmol

Explanation:

∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol

∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol

∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol

∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol

⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)

⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol

⇒ S°m,298K = 85.184 J/Kmol

4 0
2 years ago
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