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2 years ago

Chemical formurla for Lithium Phsophide

Chemistry

1 answer:

Nina [5.8K]2 years ago

3 0

Answer:

Li3P

Explanation:

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If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get:

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

3 0

3 years ago

Please show the work :(What is the estimated force applied to the box if the acceleration is .40 m/s2?

lara31 [8.8K]

Answer:

20N

Explanation:

Given parameters:

Force(N) Acceleration(m/s²)

10 0.2

? 0.4

Unknown:

The force applied when the acceleration is 0.4m/s²

Solution:

From newton's second law of motion;

Force = mass x acceleration

Since we are using the same box, let us find the mass of the box;

Force = mass x acceleration

10 = mass x 0.2

mass = $\frac{10}{0.2}$ = 50kg

Now,

The force in the second instance will be;

Force = 50 x 0.4 = 20N

8 0

2 years ago