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sleet_krkn [62]
3 years ago
12

Calculate the volume of 6 M acetic acid needed to prepare 100 mL of a 0.10 M acetic acid (CH3COOH) solution.

Chemistry
1 answer:
Zina [86]3 years ago
8 0

Answer:

V = 1.66mL acetic acid

Explanation:

dilution formula:

V1*C1 =V2*C2

⇒ V1*(6M) = (100mL)*(0.1M)

⇒ V1 =( (100mL) * (0.1M) ) / (6M)

⇒ V1 = 1.66mL acetic acid

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Ostrovityanka [42]

(D)- Sea breezes are local winds that blow from an ocean or a lake

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7 0
3 years ago
how much heat is released when 70.9g of water at 66C cools to 25C? The specific heat of water is 1 cal/gC
Serga [27]

Answer:

-12162.47 joules (or -12000 joules when accounting for significant figures)

Explanation (btw I used 1 cal as 4.184 joules because SI units are better):

q = m c delta T

q = (70.9) (4.184) (25 - 66)  

q = (70.9) (4.184) (-41)

q = -12162.47 joules

7 0
3 years ago
Which of these elements is this group?
sergey [27]

Answer:

Strontium

Explanation:

In the periodic table, an element with two (2) valence electrons is found on group 2. Group 2 is a group of the periodic table that harbors element called ALKALINE EARTH METALS. As the name implies, they are metals that possess shiny and solid characteristics at room temperature.

Group 2 elements include the following: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), barium (Ba), and radium (Ra). Based on the descriptive information in this question, the element being described is a GROUP 2 element. Based on the elements in the option, only STRONTIUM (Sr) is a group 2 element.

3 0
2 years ago
What is the molarity of a 9.0 L solution that contains 0.500 mol HCl?
victus00 [196]

Answer: 0.056

Explanation:

6 0
3 years ago
a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on
OleMash [197]

Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

(2.95L/298K ) . T2 = 2.25L

T2 = 2.25L . 298K / 2.95L

T2 = 227.2K

T°K - 273 = T°C

227.2K - 273 = -45.8°C

3 0
3 years ago
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