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3 years ago

If I add 4 spoons of sugar in one cup of water and add 2 spoons of sugar in another cup of water which one has more density?

Chemistry

1 answer:

OLEGan [10]3 years ago

4 0

The one with 4 spoons of sugar has more density.

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19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Alecsey [184]

Answer:

$m_{H_2O}=3.384gH_2O$

Explanation:

Hello,

In this case, the chemical reaction is:

$Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O$

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

$n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O$

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

$m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O$

Regards.

4 0

3 years ago

Ort

ivanzaharov [21]

Answer:

1 ethanol is right answer

Explanation:

CH3- CH2-OH

4 0

3 years ago

Read 2 more answers

Using standard free energy of formation values given in the introduction, calculate the equilibrium constant kp of the reaction

WINSTONCH [101]

d G = 2 (dG NOCl) - 2 (dG NO)

= 2 * 66.08 - 2 * 87.6 = -43.04 kJ /mol

dG = - 2.303 RT log K

-43.04 = - 2.303 RT log K

7.54 = log K

K = 3.42 * 10^7

Kp = k (RT)^dn

d n = 2 - 1 -2 = -1

Kp = (3.42 * 10^7) (8.314 * 298)^-1

Kp = 1.31 * 10^4

6 0

3 years ago

A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.

lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

$(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g$

The produced energy will be:

$=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K$

$=450.35\times 3.2$

$=1441.12 \ J$

The reaction will be:

⇒ $HCl+NaOH \rightarrow NaCl+H_{2}O$

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.

Therefore, the moles generated by NaCl will indeed be:

= $(\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}$

= $0.0318\times 0.500$

= $0.0159 \ mole \ of \ NaCl$

Now,

= $\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}$

= $906364.7$

= $90.6 \ KJ/mol \ NaCl$

7 0

3 years ago

Which of the following is a compound? H2 Ne CO2 O2

IRINA_888 [86]

Answer:

CO2

Explanation:

4 0

3 years ago