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Mrrafil [7]
3 years ago
14

85g=kg? Unit conversion.

Chemistry
1 answer:
Paraphin [41]3 years ago
3 0

Answer:

0.085 kg

Explanation:

1 g=0.001 kg  OR 1 kg=1000 g

 so \frac{85}{1000}=0.085 kg

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What is the pH of a 75.0 mL solution that is 0.045 M in weak base and 0.053 M in the conjugate weak acid
noname [10]

Answer:

7.07

Explanation:

HA = weak acid = 0.053

A+ = conjugate base = 0.045

Ka = 7.2x10^-8

Ka = [H+][A-]/HA

7 2x10^-8 = [H+][0.045]/0.053

[H+] = 7.2x10^-8 x 0.053/0.045

= 8.48x10^-8

PH = -log[H+]

= -log[8.48x10^-8]

PH = -[login.48 + log10^-8]

PH = -0.928 - (-8)log10

= 7.07

3 0
3 years ago
using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius​
lukranit [14]

Answer:

Please, see attached two figures:

  • The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

  • The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

  • Answer: <u>5.5g</u>

Explanation:

The red  arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

            10mL\times 1.0g/mL=10g

Thus, the solutibily is:

      10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl

5 0
3 years ago
If water and oil are combine If water and oil are combined in a container, the resulting liquid is a(n) A. solution. B. mixture.
mart [117]
B. Mixture is the answer
8 0
3 years ago
Who is good with science?
MrMuchimi
I can help with science.
7 0
3 years ago
A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to
Drupady [299]

Answer : The final temperature of the mixture is 29.6^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 iron = 0.499J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron = 39.9 g

m_2 = mass of water  = Density\times Volume=1g/mL\times 50.0mL=50.0g

T_f = final temperature of mixture = ?

T_1 = initial temperature of iron = 78.1^oC

T_2 = initial temperature of water = 25.0^oC

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

(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC

T_f=29.6^oC

Therefore, the final temperature of the mixture is 29.6^oC

8 0
3 years ago
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