What type of process occurs when solidified wax is converted to liquid wax on heating?

The _________ indicates that 70% of the Sun's radiation is contained on Earth and 30% is reflected back into space.

vichka [17]

i think it would be Earth's albedo hope this helps

4 0

3 years ago

A solid, pure substance containing just carbon and hydrogen is found to contain 35.8g of carbon and 3.72 g of hydrogen. how many

Igoryamba

Answer:

4.36 g of Carbon

Solution:

Step 1: Calculate the %age of Carbon in given Solid as;

Mass of Carbon = 35.8 g

Mass of Hydrogen = 3.72

Total Mass = 35.8 g + 3.72 = 39.52 g

%age of Carbon = (35.8 g ÷ 39.52 g) × 100

%age of carbon = 90.58 %

Step 2: Calculate grams of Carbon in 4.82 g of given solid as;

Mass of Carbon = 4.82 g × (90.58 ÷ 100)

Mass of Carbon = 4.36 g

3 0

3 years ago

How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5 please clear explain

11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

mg NaCl:

5g * (1000mg / 1g) = 5000mg

L Solution:

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$