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steposvetlana [31]
3 years ago
13

An element's atomic number is 65. How many protons would an atom of this element have?

Chemistry
1 answer:
KIM [24]3 years ago
7 0
65. The atomic number is the number of protons in the atom.
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A sample of gas is held at 100oc at a volume of 20 L.if the volume is increased to 40 L what is the new temperature of the gas i
ElenaW [278]

Answer:

470 °C  

Explanation:

This looks like a case where we can use Charles’ Law:  

\dfrac{V_{1}}{T_{1}} =\dfrac{V_{2}}{T_{2}}

Data:

V₁ = 20 L; T₁ = 100 °C

V₂ = 40 L; T₂ = ?  

Calculations:

(a) Convert the temperature to kelvins

T₁ = (100 + 273.15) K = 373.15 K

(b) Calculate the new temperature

\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{20 L}}{\text{373.15 K}} &=&\dfrac{\text{40 L}}{T_{2}}\\\\{\text{15 000 K}} & = & 20T_{2}\\T_{2} & = &\dfrac{\text{15 000 K}}{20 }\\\\T_{2} & = & \textbf{750 K}\\\end{array}

Note: The answer can have only two significant figures because that is all you gave for the volumes.

(c) Convert the temperature to Celsius

T₂ = (750 – 273.15) °C = 470 °C

3 0
3 years ago
How many grams of sodium benzoate, C H CO Na, have to be added to 1.50 L of a 0.0200 M solution of benzoic acid, C H CO H, to ma
Dmitrij [34]

Answer: 2.8275grams

Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid

C6H5COOH....> C6H5COO- + H+

Ka = [C6H5COO-] [H+]/ [C6H5COOH].......(1)

using Ka = 6.5× 10^-5, [C6H5COOH] = 0.02M. PH= - log[H+] ....> [H+]= 10^-4M.

Putting the values in(1)

[C6H5COO-]= 6.5× 10^-5 × 0.02/ 10^-4

[C6H5COO-] = 0.013M = Molarity of sodium benzoate

Mole(C6H5COONa) = 0.013 × Volume = 0.013mol/litre × 1.5 litre

Mole(C6H5COONa) = 0.0195mol

Mass(C6H5COONa) = 0.0195 × Molar mass

Mass(C6H5COONa) = 2.8275g

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