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ryzh [129]
3 years ago
6

How many oxide ions are in 0.55 grams of titanium oxide?

Chemistry
1 answer:
attashe74 [19]3 years ago
3 0

Answer: There are 0.052\times 10^{23} oxide ions

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{0.55gg}{64g/mol}=0.0086moles

1 mole of TiO contains = 6.023\times 10^{23} oxide ions

Thus 0.0086 moles of TiO contains = \frac{6.023\times 10^{23}}{1}\times 0.0086=0.052\times 10^{23} oxide ions

There are 0.052\times 10^{23} oxide ions

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2 years ago
How many pounds of ice are required to absorb 4900 kJ of heat as the ice melts? The heat of fusion of water is 0.334 kJ/g.
Goshia [24]

Answer:

m = 32.34 pounds of ice.

Explanation:

In this case we need to use the following expression of heat:

q = m * ΔHf   (1)

Where:

q: heat absorbed in J or kJ

m: mass of the compound in g

ΔHf: heat of fusion of the water in kJ/g

We are asked to look for the mass of ice in pounds, so after we get the grams, we need to convert the grams to pounds, using the following conversion:

1 pound --------> 453.59 g   (2)

So, we have the heat and heat of fusion, from (1) let's solve for the mass, and then, using (2) the conversion to pounds:

q = m * ΔHf  

m =  q / ΔHf

m = 4900 / 0.334 = 14,670.66 g of ice

Now, the conversion to pounds:

m = 14,670.66 g * 1 pound/453.59 g

<h2>m = 32.34 pounds of ice.</h2>

Hope this helps  

5 0
3 years ago
1. power utilizes moving water to generate electricity.
Marianna [84]
Hydroelectric energy
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Which compound is made from metals and non metals​
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Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

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3 years ago
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