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QveST [7]
3 years ago
15

How many moles of cesium (Cs) atoms are in 675g Cs?

Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
8 0
The atomic mass of Cs is 132.9. So the mole number of 675 g cesium is 675/132.9 = 5.08 mole. Then the answer should be 5.08 mole.
timofeeve [1]3 years ago
3 0

Answer: 5.08 moles

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 number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

given mass of Cesium (Cs) = 675 g

Molar mass of Cesium (Cs) = 132.90 g/mol

Putting in the values we get:

\text{Number of moles}=\frac{675g}{132.90g/mol}=5.08moles

Thus there are 5.08 moles of cesium in 675 g of cesium.

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What must be true of a disproportionate substance?
Rudik [331]

Answer:

Answer is D.it gains and loses electrons.

Explanation:

I hope it's helpful!

4 0
3 years ago
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A rigid 3.80 L sealed vessel contains 0.650 mol Ne, 0.321 mol Kr, and 0.190 mol Xe. Find the density of the mixture in g/L.
Eddi Din [679]

Answer:

17.09g/L

Explanation:

Density = total mass of elements/ volume

We need to find the mass of each mixture constituents using their molar mass:

mole = mass/molar mass

For Neon (Ne) which contains 0.650mol;

0.650 = mass/20.18

mass = 0.650 × 20.18

mass = 13.12g

For Krypton (Kr) which contains 0.321mol;

0.321 = mass/83.79

mass = 0.321 × 83.79

mass = 26.89g

For Xenon (Xe) which contains 0.190mol;

0.190 = mass/131.3

mass = 0.190 × 131.3

mass = 24.95g

Total mass = 13.12g + 26.89g + 24.95g = 64.96g

Density = total mass / volume

Density = 64.96g / 3.80L

Density of the mixture = 17.09g/L

7 0
3 years ago
Balance this chemical equation.
Vadim26 [7]

Answer:

Explanation:

Reaction Given

    NaHCO₃ (s) + HC₂H₃O₂ (aq) -------> CO₂ (g) + H₂O (l) + NaC₂H₃O₂ (aq)

Balance Equation = ?

Solution:

Balance Chemical Equation:

A balanced chemical equation is that in which the number of the reactant atoms equal to the number of the product atom.

For example if the carbon at the reactant side is 3, the number of carbon must be 3 on the product side.

Check the chemical equation and count the number of atoms at the reactant side and product side.

If the number of atoms of the reactants equal to the number of atoms of products then the reaction is balanced but if not equal then reaction is not balanced

Balancing is an trial and error process to get a balance reaction.

First we count the number of atoms of reactant and product

Number of atoms of Reactant:

Na = 1

O = 5

H = 5

C = 3

Number of atoms of Product:

Na = 1

O = 5

H = 5

C = 3

So this a balance equation as the atoms of the reactant atom equal to the number of the product atoms.

Na atom is 1 on both reactant and product side

Oxygen atoms are 5 on reactant and 5 on product side

hydrogen atoms are 5 on reactant and 5 on product side

Carbon atoms are 3 on reactant and 3 on product side

So all the compounds in the reaction have 1 as its coefficient in the blanks

<u />

<u>1   </u>NaHCO₃ (s) +  <u>1  </u>HC₂H₃O₂ (aq) -----> <u>1  </u>CO₂ (g) + <u>1  </u>H₂O (l) + <u>1  Na</u>C₂H₃O₂ (aq)

7 0
3 years ago
These two elements are most likely good conductors of heat and electricity and shiny when solid.
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Round the following number to four significant figures and express the result in standard exponential notation: 442,722 *
OleMash [197]

Answer:

4.427 × 10⁵

Explanation:

In mathematics, rounding up a number is a way to get a more simpler figure but not too distant from the original number. Rounding up a whole number like this (442722) to significant figures means having only four non-zero digits in the figure i.e. 442722 ~ 442700.

This means that 442722 has been rounded up to 4s.f. Hence, to use scientific notations to represent our answer, we count the number of decimal places from right to left i.e.

442700 = 5 decimal places if the point is between 4 and 4 i.e. 4.427

Therefore, the figure 442700 can be written as 4.427 × 10⁵ using scientific notation.

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