All categories

2 years ago

PLEASE HELP?? :| AS SOON AS POSSIBLE!!

Chemistry

2 answers:

frutty [35]2 years ago

6 0

B hope this helps, it sounds right

Kitty [74]2 years ago

5 0

Oceans cover most of the planets surface and contain nearly all its water.

You might be interested in

Help me with homework.

andrey2020 [161]

The chemical equation is said to be balanced if the number of atoms in the reactants and products is the same

<h3>Further explanation</h3>

Equation balanced ⇒ total number of atoms in reactants(on the left)= total number of atoms in products(on the right)

H₂+O₂---> H₂O

Reactants : H₂, O₂

Products : H₂O

not balanced

H₂O₂ ---> H₂O+O₂

Reactants : H₂O₂

Products : H₂O, O₂

not balanced

Na+O₂ ---> Na₂O

Reactants : Na, O₂

Products : Na₂O

not balanced

N₂+H₂ ---> NH₃

Reactants : N₂, H₂

Products : NH₃

not balanced

P₄+O₂---> P₄O₁₀

Reactants : P₄, O₂

Products : P₄O₁₀

not balanced

Fe+H₂O ----> Fe₃O₄ + H₂

Reactants : Fe, H₂O

Products : Fe₃O₄

not balanced

7 0

3 years ago

How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

gulaghasi [49]

Answer:

$\boxed {\boxed {\sf 10.2 \ kJ}}$

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

$q=mc \Delta T$

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

m= 225 g
c= 0.897 J/g° C
ΔT= 50.5 °C

$q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

Multiply the first two numbers. The units of grams cancel.

$q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

$q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)$

$q= (201.825\ J / \textdegree C)(50.5 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 201.825 \ J * 50.5$

$q= 10192.1625 \ J$

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

$\frac { 1 \ kJ}{ 1000 \ J}$

Multiply by the answer we found in Joules.

$10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}$

$10192.1625 * \frac{ 1 \ kJ}{ 1000 }$

$\frac {10192. 1625}{1000} \ kJ$

$10.1921625 \ kJ$

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

$10.2 \ kJ$