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klemol [59]
2 years ago
7

HELP PLEASE!!!!!!!!! What if you started with 100 grams of wood and only have 95 grams of ashes produced, is mass still conserve

d? If so where is the missing 5 grams?
Chemistry
1 answer:
Lerok [7]2 years ago
7 0
My guess is smoke. Smoke is matter.
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Help quick! Will mark brainlest!
Ivenika [448]

Answer: 10HSiCl3 + 15H2O → H10Si10O15 + 30HCl

Explanation:

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2 years ago
How many significant figures does 4,982cm have
Elodia [21]

Answer:

4 significant figures.

Explanation:

All non-zero numbers are significant figures.

8 0
3 years ago
I need help !!!!!!!!!!! Plz
pishuonlain [190]

Answer: 1: B. It will speed up the reaction.

2: A. Increase the concentration of a reactant.

3: A. They act as catalysts.

4: C. A catalyst decreases the activation energy needed for a reaction.

5: D. collision theory.

Explanation: I just took the quiz:)

5 0
3 years ago
Which of the following elements is a metalloid
klemol [59]
A metalloid can be:
 - Boron (B)
 - Silicon (Si)
 - Germanium ( Ge)
 - Arsenic (As) 
 - Antimony ( Sb)
 - Tellurium (Te)
 - Polonium (Po)

Hope this helps :)
6 0
3 years ago
Read 2 more answers
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
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