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2 years ago

Label each description as ionic, covalent, or both.

2 answers:

8 0

Atoms share pairs of electrons: covalent
Atoms transfer electrons: ionic
Atoms have electrostatic attraction:
Atoms bond together:Both

klemol [59]2 years ago

4 0

Answer :

Atoms share pairs of electrons : Covalent

Atoms transfer electrons : Ionic

Atoms have electrostatic attraction : Ionic

Atoms bond together : Both

Explanation :

Covalent compound : The compounds where the atoms are covalently bonded.

Covalent bonds are formed by the equal sharing of electrons. These bonds are commonly formed between two non-metals.

Ionic compound : The compounds where the atoms are bonded through ionic bond and atoms are bonded by the electrotstatic attraction.

Ionic bonds are formed by the complete transfer of electrons. The atom which looses the electron is considered as an electropositive atom and the atom which gains the electron is considered as electronegative atom. These bonds are formed between one metal and one non-metal.

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100. cal of heat are added to 18.0 g of ethanol (0.581 cal/g °C) originally at 23 °C. The final temperature is ____________.

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Answer:

Final temperature is 32.56 °C

Explanation:

The specific heat of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1°C.

The following equation/formula is used;

Q = m × c × ΔT

Where; Q= amount of heat supplied

(cal)

M= mass of ethanol (g)

C= specific heat of ethanol

(cal/g °C)

ΔT= change in temperature (°C)

i.e. (final temperature - initial

temperature)

According to the question, Q= 100 calories (cal), M= 18g, C= 0.581 cal/g °C, initial temperature = 23°C, final temperature = ?

Hence, we insert our values into the equation;

Q = m × c × ΔT

ΔT = Q/mc

(Final T - Initial T) = Q/mc

(Final T - 23) = 100/ 18 × 0.581

(Final T - 23) = 100/10.458

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2 years ago

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Answer: The mass of the gas is 18.3 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

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$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}$

$1.55=\sqrt{\frac{44}{M_{X}}$

Squaring both sides and solving for $M_{X}$

$M_{X}=18.3g/mol$

Hence, the molar mas of unknown gas is 18.3 g/mol.

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