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3 years ago

Energy is just like any other chemical or physical process, where it cannot be created or destroyed. Which Law states this fact?

Chemistry

1 answer:

Studentka2010 [4]3 years ago

3 0

The answer should be law of conservation of energy

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What are the products of the neutralization reaction between HCl and

liraira [26]

Answer:

opinion b

Explanation:

the product of neutralization reaction between hcl and CA(oh)2 is option b.

4 0

3 years ago

Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at

WITCHER [35]

<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

$0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL$

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

$\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL$

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0

3 years ago

B. What is the mass of a sample of mercury if its volume is 4.35 cubic centimeters?

icang [17]

Answer:

Mercury has a density of 13.6g/mL

6 0

3 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago

11.You and a friend find a rusty wire coat hanger near the

gtnhenbr [62]

Answer:

because he would've not known properly about it!

7 0

3 years ago