Here we have to calculate the heat required to raise the temperature of water from 85.0 ⁰F to 50.4 ⁰F.
10.857 kJ heat will be needed to raise the temperature from 50.4 ⁰F to 85.0 ⁰F
The amount of heat required to raise the temperature can be obtained from the equation H = m×s×(t₂-t₁).
Where H = Heat, s =specific gravity = 4.184 J/g.⁰C, m = mass = 135.0 g, t₁ (initial temperature) = 50.4 ⁰F or 10.222 ⁰C and t₂ (final temperature) = 85.0⁰F or 29.444 ⁰C.
On plugging the values we get:
H = 135.0 g × 4.184 J/g.⁰C×(29.444 - 10.222) ⁰C
Or, H = 10857.354 J or 10.857 kJ.
Thus 10857.354 J or 10.857 kJ heat will be needed to raise the temperature.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
<h3>What is radioactivity?</h3>
Radioactivity is the spontaneous decay of a substance with emission of radiation.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.
Learn more about radioactivity at: brainly.com/question/3603596
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It forms <span>calcium phosphate and potassium nitrate
</span>2 K3PO4 + 3Ca(NO3)2 --> Ca3(PO4)2 + 6KNO3
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
Just search up how much the weight is equal to another, then multiply it, thats what i do lol sorry if im no help
