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2 years ago

What is the chemical equation for foam fight

Chemistry

1 answer:

Soloha48 [4]2 years ago

8 0

Answer:

Hydrogen Peroxide

Explanation:

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Some animals that live in extremely cold environments protect themselves from freezing by increasing the molality of certain che

lara [203]

Molality is defined as the number of moles of solute dissolved per 1000g(1kg) of solvent.
It is represented by m

$\\ \sf \longmapsto m=\dfrac{Moles\:of\:solute}{Wt\:of\:solvent(in\:g)}\times 1000$

8 0

3 years ago

Can animals detect radio waves

Svetach [21]

Answer:

No.

Explanation:

No organism can detect X-rays or radio waves

4 0

2 years ago

Read 2 more answers

Please answer my questions. I really need them soon. I will post the links for other questions in the comments.

Assoli18 [71]

Explanation:

option option B is the correct answer of given statement helium-4(He)=2

6 0

2 years ago

Question<br> What is the molarity of a 400 mL solution containing 0.60 moles of NaCl?

Mumz [18]

Answer:

0.24 M

Explanation:

Molarity = Moles solute / Liters solution

Step 1: Identify variables

400 mL = Liters solution

0.60 moles = Moles solute

Step 2: Identify conversions

1 L = 1000 mL

Step 3: Convert mL to L

400mL(1 L/1000mL) = 0.4 L

Step 4: Find molarity

M = (0.4 L)(0.60 mol) = 0.24 M

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