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3 years ago

The lattice structure in a metal is _____ than the lattice structure of an ionic compound.

1 answer:

8 0

The lattice structure in a metal is weaker than the lattice structure of an ionic compound. This is because the metals have free electrons which can freely move around while ionic compounds are strongly bonded. Hope this answers the question.

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Are atoms the smallest particles we know about, or are there smaller ones?

Schach [20]

Answer:

That we know about yes, but physicians still say there are more things out there. (we do not know about these yet tho)

Explanation:

7 0

3 years ago

A helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 20.0 °C is released. What volume will the ballo

meriva

Answer:

The balloon will occupy a volume of 6.48 L

Explanation:

Step 1: Data given

internal pressure = 1.00 atm

volume = 4.50 L

Temperature = 20.0 °C

Step 2: Calculate new volume via the ideal gas law

P*V = n*R*T

(P1*V1)/ T1 = (P2*V2)/T2

⇒ with P1 = 1.00 atm

⇒ with V1 = 4.50 L

⇒ with T1 = 20.0 °C = 293 Kelvin

⇒ with P2 = 0.600 atm

⇒ with V2 = TO BE DETERMINED

⇒ with T2 = -20°C = 253 Kelvin

(1.00atm * 4.50 L)/293 Kelvin = (0.600 atm*V2) / 253 Kelvin

0.01536 = 0.00237 V2

V2 = 6.48 L

The balloon will occupy a volume of 6.48 L

7 0

2 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$