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3 years ago

What type of science is chemistry?

Chemistry

2 answers:

d1i1m1o1n [39]3 years ago

6 0

Answer:

It is learning about chemical reactions. In chemistry you learn about the elements of the periodic table and how reactive they are.

Explanation:

PLEASE MARK ME THE BRAINLIEST :D!!

KonstantinChe [14]3 years ago

6 0

Answer:

physical science

Explanation:

because it touches all other natural sciences. biology,physics, etc. it is known as the central science:)

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0

3 years ago

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

Citrus2011 [14]

Depression in freezing point (Δ $T_{f}$ ) = $K_{f}$ ×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$ ,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$ )

Thus, (Δ $T_{f}$ ) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ $T_{f}$ ) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ $T_{f}$ ) = - $0.03162^{0}C$

8 0

3 years ago

Read 2 more answers

The half-life of a strontium isotope is 25 years. How much of this particular isotope will remain after 100 years if the startin

Phoenix [80]

Answer:

1 gram

Explanation:

Half life = 25 years

Starting mass = 16 grams

Time = 100 years

Number of half lives = Time / Duration of Half life = 100 / 25 = 4

After first Half life;

Remaining mass = 16 / 2 = 8 g

After Second Half life;

Remaining mass = 8 / 2 = 4 g

After Third Half life;

Remaining mass = 4 / 2 = 2 g

After Fourth Half life;

Remaining mass = 2 / 2 = 1 g

3 0

2 years ago

Simple machines make work easier. Most simple machines reduce the amount of effort force needed to move an object. This cart has

Kruka [31]

I'm pretty sure it's D

8 0

3 years ago

Analyze the role coefficients in a chemical reaction play in stoichiometry

kumpel [21]

Answer:

This is known as the coefficient factor

Explanation:The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element.

7 0

2 years ago