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3 years ago

How do the strong ionic bonds affect the boiling and melting points of salts

Chemistry

1 answer:

likoan [24]3 years ago

3 0

Answer:

The more energy needed, the higher the melting point or boiling point . Since the electrostatic forces of attraction between oppositely charged ions are strong, their melting and boiling points are high.

Explanation:

hope this helps.

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The Resource Conservation and Recovery Act controls hazardous waste from its creation to its disposal. Please select the best an

o-na [289]

Answer:

True

Explanation:

Resource Conservation and Recovery Act (RCRA). 1976 of United States Environmental Protection Agency(EPA) empowers EPA to control the production, transportation, storage, treatment and disposal of hazardous waste. The RCRA act was amended in 1984 and 1986 to include Waste minimization along with appropriate disposal (not in the landfill site) and tackling of petroleum hazardous waste respectively along with other waste.

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3 years ago

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According to the octet rule, which of elements will have a tendency to loss 2 electrons?

Aneli [31]

The correct option is STRONTIUM.
Strontium is a group 2 element, that means it has two electrons in its outermost shell. This element will prefer to lose these two electrons in its outermost shell in order to attain the octet form, therefore, it will form electrovalent bond with non metals which it can donate two electrons to.

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2 years ago

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$