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nevsk [136]
3 years ago
5

Identify the anode and cathode please

Chemistry
1 answer:
Nuetrik [128]3 years ago
6 0

Answer:

anode is negative. cathode is positive.

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What is the preasure in atmospheres of 20 mol of nitrogen gas in 36.2 L cylinder at 25 degrees C?
Savatey [412]

Answer:

P = 13.5 atm

Explanation:

Given that

No. of moles, n = 20 moles

Volume of nitrogen gas = 36.2 L

Temperature = 25°C = 298 K

We need to find the pressure of the gas. Using the ideal gas equation

PV = nRT

Where

R is gas constant, R=0.082057\ L-atm/K-mol

So,

P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm

so, the pressure of the gas is equal to 13.5 atm.

7 0
2 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

8 0
1 year ago
Why do atoms tend to be bonded to other atoms?
Aleks04 [339]

Answer:

Atoms are often more stable when bonded to other atoms

Explanation:

Like for example let's say ionic bonds..... Since one atom has to lose specific electrons to be stable and the other needs the electrons from the other atom to be stable.....

3 0
3 years ago
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The solid form of a substance is usually more dense than its liquid and gaseous forms. Similarly the liquid form is usually more
solniwko [45]
The solid form of a substance is usually more dense than its liquid and gaseous forms. Similarly the liquid form is usually more dense than the gaseous form. Ice floating in water is an exception that breaks the general density rule. So option “A” is the correct option in regards to the given question. In case of ice formation, actually the density of water decreases by about 9%. This is the main reason behind ice floating in water. Pure water has the maximum density at 4 degree centigrade.


7 0
3 years ago
SEP Engage in Argument A classmate claims that state of matter is an extensive property because it can vary with temperature. Fo
Andrews [41]

Answer: I agree with the student because in the question prior to this One question stated thatExtensive properties very with the amount of matter ina sample, so yes i agree.

Explanation:

4 0
3 years ago
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