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2 years ago

Please help find out code?

Chemistry

1 answer:

Nat2105 [25]2 years ago

4 0

You might need to take more pictures so we can see all the equations clearly

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How is the density of an object in kg/L related to its density in g/mL? (1 point)

umka2103 [35]

Answer:

kg =1000g ml =1/1000 L

Explanation:

1000g/L=1g/1/1000L=1000g/L

The same

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3 years ago

What is a Product? ASAP

Masja [62]

Answer:

Do you want me to give you the definition in math, science or just what a product means because there's different types.

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3 years ago

The reaction below is at dynamic equilibrium. mc004-1.jpg Which statement is true for the equilibrium system? The concentration

enyata [817]

When a system is in dynamic equilibrium, the forward reaction rate and the backward reaction rate are equal or occurs at the same rate. Therefore, the third option above is the most accurate one. Hope this answers the question. Have a nice day.

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3 years ago

Read 2 more answers

What's Meso compounds? Please explain it to me with an appropriate example. I'll mark you as Brainliest

slamgirl [31]

Answer:

this can be explained as a non optically active member of a set of stereoisomer, at least two of which are optically active

Explanation:

this means that despite containing two or sterogeonic center the molecules are not chiral

example:

the example above shows at least two choral center and internal mirror plane

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3 years ago

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

7 0

3 years ago