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PilotLPTM [1.2K]
4 years ago
14

Pls help asap...

Chemistry
1 answer:
Dmitriy789 [7]4 years ago
7 0
The answer to this question would be <span>repeatable and verifiable. If the result of the research can't be repeated then you can't deduce anything from it. The result should be verified so no one can falsify a claim of observation.

Morality and the inexpensive shouldn't be mandatory.</span>
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135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5 C to 100.0 C what is the specific heat of this meta
Harman [31]
The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5. 
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Digiron [165]

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8 0
3 years ago
When an acid solution exactly neutralizes a base
o-na [289]

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Explanation:

7 0
3 years ago
An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?
ra1l [238]
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant 
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT 
where 
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation 

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL 
therefore answer is 
c. 448 mL 
5 0
3 years ago
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