All categories

2 years ago

Why is it dangerous to have electric appliances near you when you are taking a bath?

Chemistry

1 answer:

VashaNatasha [74]2 years ago

4 0

Answer:

Because water conducts electricity, so if an electrical current were to meet the water while you're in there taking a bath, you would essentially be surrounded by a huge conductor.

You might be interested in

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at

pochemuha

Total internal energy change is equals to -44.83kJ

Q=-73.2kJ (negative sign indicates that heat was released by the system),

P= 50.0atm

ΔU= Q + W, FIRST LAW OF THERMODYNAMICS..........(1)

ΔV= Final volume - initial volume= 2.00 litre - 7.60litre= -5.60litre

work done by the system (w)= -PΔV

w= -(50.0×(-5.60)) atm×litre= 280atm litre

1 atm litre= 101.325J

w= 280 ×101.325 J= 28,371J

1kJ=1000J,

w=28.37KJ,

so putting in the values in equation (1)...

energy change(ΔU) = -73.2 kJ + 28.37 kJ

= - 44.83 kJ

7 0

2 years ago

What will litmus paper turn when it reacts with sodium nitrate?

castortr0y [4]

A solution of sodium nitrate will get make litmus paper wet. A solution of NaNO3 is neutral. Therefore, there is no change in color.

6 0

3 years ago

Give the formula of a compound in unpolluted air

77julia77 [94]

Carbon dioxide and oxigen

6 0

3 years ago

EXTRA CREDIT

marin [14]

Answer:

$\huge\boxed{\sf 36\ H\ atoms}$

Explanation:

Molecular formula from Glucose:

C₆H₁₂O₆

3 moles of Glucose:

3C₆H₁₂O₆

In 1 mole of Glucose, there are 12 hydrogen atoms.

In 3 moles:

= 12 × 3

= 36 H atoms

$\rule[225]{225}{2}$

5 0

2 years ago

An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .

grandymaker [24]

Answer: The specific heat of the unknown metal is $0.897J/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $5040$ Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i=(64.7)^0C$

Putting in the values, we get:

$5040=86.8\times c\times 64.7^0C$

$c=0.897J/g^0C$

The specific heat of the unknown metal is $0.897J/g^0C$

4 0

3 years ago