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3 years ago

Sugar dissolving in warm water is chemical or phyiscal change

1 answer:

5 0

Hey there! Great question;) Answer:Physical change Explanation: When sugar mixes with water, at the end, the chemical formulas are the same. Nothing has changed! I hope this helps;)

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If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?

solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂ = 4.9kg = 4900g

Unknown:

Number of molecules = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CO₂ = 12 + 2(16) = 44g/mol

Number of moles = $\frac{4900}{44}$ = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

1 mole = 6.02 x 10²³molecules

111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules

5 0

3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

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

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

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

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

Help! How is this incorrect?

Ainat [17]

On the first one it is supposed to be 18. when you have a +1 charge you subtract it once. how i got 18 tho was from the protons. there was 19 so i subtracted that with 1 and got 18. hope that helped! :)

btw i’m not the best at explaining, i’m sorry :/

8 0

2 years ago

the atomic mass of N is 14.01 g/mol and the atomic mass of H is 1.008 g/mol. what is the molecular mass of NH3

nikdorinn [45]

Molecular mass= (14.01∗1)+(1.008∗3)
14.01+3.024=17.03g/mol

7 0

3 years ago

If the actual yield of sodium chloride from the reaction of

lions [1.4K]

If the actual yield of sodium chloride from the reaction of 8.3 g of sodium and 4.5 g of chlorine is 6.4 g, what is the percent yield?

Answer : 2Na(s) + Cl2(g) → 2NaCl(s)

Explanation: Friend me!!!

7 0

2 years ago