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Darina [25.2K]
3 years ago
15

It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium c

hloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution? The freezing-point depression constant for water is Kf = –1.86°C/m. In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.
Chemistry
1 answer:
Yuri [45]3 years ago
3 0

Answer:

Di ko po mahanap

Explanation:

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<h2>All Group 1 metals form halides that are white solids at room temperature. The melting point is correlated to the strength of intermolecular</h2>
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7 0
3 years ago
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
4 years ago
NaOH+H3PO4=Na3PO4+H20
statuscvo [17]
Wait is that suppose to be a question??!!
4 0
3 years ago
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