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3 years ago

What is an experiment?

Chemistry

1 answer:

OLEGan [10]3 years ago

6 0

Answer:

a scientific procedure undertaken to make a discovery, test a hypothesis, or demonstrate a known fact.

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When coke burns in air then evolved gas is

Arturiano [62]

Answer:

Explanation:

Carbon (coke) burns in air to form carbon dioxide gas.

(i) C(s) + O2 ↑= CO2 ↑

7 0

3 years ago

The metals in Groups 1A, 2A, and 3A ____.

Ket [755]

The correct answer is D.

I hope that helped! c:

6 0

3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

3 0

2 years ago

Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife

yuradex [85]

Answer:

$\boxed{\text{2408 min}}$

Explanation:

The integrated rate law for radioactive decay is

$\ln\dfrac{N_{0}}{N_{t}} = kt$

1. Calculate the decay constant

$\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}$

8 0

3 years ago

Ito Ang naimbento ni graham bell na nagpabilis sa pakikupagtalastasan

Feliz [49]

Answer:

ФОТОГРОФИРУЙ

ИЛИ НАПИШИ

В КОМЕНТЫ

7 0

2 years ago