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Ivan
3 years ago
7

How many grams of water can be heated 20C by the addition of 1000J?

Chemistry
1 answer:
nydimaria [60]3 years ago
4 0
I hope this is correct

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List 2 of the important properties of gases​
horrorfan [7]

Answer:

pressure, temperature and volume

Explanation:

3 0
3 years ago
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How many grams of na3po4 will be needed to produce 650. mL of a solution that has a concentration of na+ ions of 1.40 M
SashulF [63]

Answer:

I think its 1.40M Na^+ we convert to Na3PO4 which means we need 1.40M x (1 mole Na3PO4/3 mol Na^+) = 1.40 x 1/3 = 0.467 M Na3PO4.

Explanation:

7 0
3 years ago
The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, ins grams, must react to produce 50
Ivan

Answer:

56.28 g

Explanation:

First change the grams of oxygen to moles.

(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂

You have to use stoichiometry for the next part.  Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced.  Convert from moles of O₂ to moles of H₂O using this relation.

(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O

Now convert moles of H₂O to grams.

(3.125 mol) × (18.01 g/mol) = 56.28125 g

Convert to significant figures.

56.28125 ≈ 56.28

5 0
3 years ago
n a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the ________, while the antibonding
Kruka [31]

Answer:

In a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the valence band, while the antibonding orbitals that are completely empty are referred to as the conduction band.

The conduction band occupies a higher energy level than the valence band. The band gap is what separates the two orbitals.

3 0
2 years ago
When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.
sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 15.8J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m = mass of water = 100.0 g

\Delta T = change in temperature = T_2-T_1=(32.0-23.9)=8.1^oC

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

q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]

q=3513.8J=3.5138kJ        (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole

Now,

\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

4 0
3 years ago
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