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AnnyKZ [126]
3 years ago
5

Explain why lithium fluoride is a solid at room temperature

Chemistry
1 answer:
SashulF [63]3 years ago
8 0
Lithium fluoride is a solid at room temperature because it is a salt that is held together by ionic bonds. Lithium. fluoride has a giant ionic structure.
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How are the sizes of stars measured?<br><br><br> I just need a simple answer
S_A_V [24]
They are determined by the length of time it takes for light to travel from the Earth to the desired star
7 0
2 years ago
This particle determines what element you have - the element's identity.
scoundrel [369]

Answer:

protons

These particles do not vary in quality from one element to another; rather, what gives an element its distinctive identification is the quantity of its protons, called its atomic number. Protons and neutrons contribute nearly all of an atom's mass; the number of protons and neutrons is an element's mass number

Explanation:

4 0
3 years ago
How many grams are in 2.3 moles of calcium phosphate, Ca3(PO3)2
irina1246 [14]
Moles = mass/molar mass
moles = 2.3
molar mass = 278
=> mass = moles*molar mass = 639.4g
6 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
How much heat is required, in calories, to raise the temperature of 57.8 g of silver from
Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

8 0
2 years ago
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