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ss7ja [257]
3 years ago
7

Can a liquid substance diffuse in another liquid substance

Chemistry
1 answer:
Zanzabum3 years ago
8 0
<span>Yes, a liquid substance diffuse in another liquid substance. Rate of diffusion depends on liquid's viscosity temperature and liquid.

</span>
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The chemical symbol for sodium is <br> a. na. <br> b. n. <br> c. s. <br> d. sn.
gayaneshka [121]
The chemical symbol for sodium is Na
8 0
3 years ago
The molar mass of silver (Ag) is 107.87 g/mol.
Setler79 [48]

Answer:

3.53 g

Explanation:

To convert from atoms to moles, you need to know Avogadro's number.  Avogadro's number (6.022 × 10²³) is how many atoms there are in one mole of a substance.  Use this to convert.

(1.97 × 10²² atoms) ÷ (6.022 × 10²³ atoms/mol) = 0.0327 mol

Now that you have moles, use the molar mass to convert to grams.

(0.0327 mol) × (107.87 g/mol) = 3.53 g

4 0
3 years ago
Why is fractional distillation not used to produce drinking water from sea water?
Natasha_Volkova [10]

Answer:

Please mark me brainliest. the answer is

Explanation:

The problem is that the desalination of water requires a lot of energy. Salt dissolves very easily in water, forming strong chemical bonds, and those bonds are difficult to break. Energy and the technology to desalinate water are both expensive, and this means that desalinating water can be pretty costly.

4 0
3 years ago
I need help with two questions.
vodomira [7]

Answer:

1) their atomic numbers

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8 0
3 years ago
Read 2 more answers
A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori
Nitella [24]

Answer : The specific heat of metal is 0.481J/g^oC.

Explanation :

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 metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of metal = 129.00 g

m_2 = mass of water = 45.00 g

T_f = final temperature = 39.6^oC

T_1 = initial temperature of metal = 97.8^oC

T_2 = initial temperature of water = 20.4^oC

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

129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC

c_1=0.481J/g^oC

Therefore, the specific heat of metal is 0.481J/g^oC.

4 0
3 years ago
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