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butalik [34]
3 years ago
15

Heeellllpppp ASAP ......

Chemistry
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

I saw B, he vented SUS

Explanation:

Iodoethane Is the answer.

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on a hot summer day, water droplets often form on the outside of a cold glass. the water droplets form when water vapor in the a
dangina [55]


this is known as condensation when cold and warm temperatures meet


7 0
4 years ago
The mass of a 50-mL volumetric flask is 40.638 g. After adding 15.0 mL of 95% ethanol and adding enough water to complete the vo
Nuetrik [128]

Answer

0.9516 grams / mL (50.00 has 4 sig digs.)

Remark

You have a couple of extraneous numbers there. You don't care about anything except the mass of the flask +  water/alcohol mixture  (88.219 grams). and the mass of the flask (40.638 grams)

Formulas

  • mass water/alcohol mixture = mass of the flask with fluid - mass flask
  • density = mass / volume

Solution

mass water/alcohol mixture = 88.219 - 40.638 = 47.581

  • Volume = 50 mL
  • Density = mass / Volume
  • Density = 47.581/50
  • Density = 0.95162  There are 4 sig digs so the answer should be
  • 0.9516
3 0
3 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
Which of the following is not a natural
Morgarella [4.7K]

Answer:

cup...................

6 0
3 years ago
Read 2 more answers
Help!!!!
V125BC [204]

Answer:D and C sorry if im wrong

Explanation:

8 0
3 years ago
Read 2 more answers
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