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Stells [14]
3 years ago
13

What will lithium and bromine make when they bond

Chemistry
2 answers:
Doss [256]3 years ago
6 0

Answer:

When lithium and bromine bond it makes lithium bromide.

Rus_ich [418]3 years ago
5 0

Answer:

they will form a compound

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A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming
ahrayia [7]

Answer:

sorry but i cant

Explanation:

8 0
3 years ago
Can anyone help me part 1
valina [46]
I can help you with part 1
8 0
3 years ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0
3 years ago
How are repetition and replication alike and different?
Nat2105 [25]
<span>Both provide approaches to confirming the result of experimentation. Repetition can be developed by one scientist or team continually achieving the expected result but replication requires an independent person or team shows thay can arrive at the same answer independently</span>
6 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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