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Nady [450]
3 years ago
15

How many significant figures are in this number? 9.678 grams

Chemistry
2 answers:
Georgia [21]3 years ago
7 0

Answer:

4 significant figures

Reason

The reason is count from 9 and all the decimal. it would be different if there was 0.09768.

KengaRu [80]3 years ago
5 0
There are 4 of them
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Malik analyzed three samples to test which element was a metalloid. The table shows his results. Which element (A, B, or C) is m
11111nata11111 [884]

the answer is c :)

have a good day (good vibes only)

- sav <3

7 0
3 years ago
Read 2 more answers
The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol

8 0
4 years ago
Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L

5 0
4 years ago
А. B<br> Which of the following is true about the picture above? What would it be?
JulijaS [17]

Answer:

Container A is a solution and Container B is a colloid.

Explanation:

Container A is a solution as the tyndall efffect is not present. You cannot see the light bean through the glass.

Container B is a suspension or a colloid as you can see the beam through the glass.

The first option proposes Container A is a colloid...this is not correct

The third option is not true as the tyndall effect is only displayed in container B

The fourth option is not correct the tyndall effect is not present in container A meaning it would not be a suspension or a colloid and you can see the tyndall effect in container B indicating its not a solution

8 0
4 years ago
According to the chemical reac- tion, if there are 4.5 mol of oxy- genated hemoglobin present in an excess of carbon monoxide, h
Deffense [45]
A single molecule of hemoglobin can bind to 4 molecules of oxygen gas. However, hemoglobin has a greater affinity for carbon monoxide than oxygen. Therefore, an excess of carbon monoxide in the presence of oxygenated hemoglobin will result in the displacement of each oxygen atom for a carbon monoxide atom.

Hb(O2)4 (aq) + 4 CO(g) --> Hb(CO)4 (aq) + 4 O2(g)

With an excess of carbon monoxide, it is safe to assume that each oxygen molecule will be displaced with a carbon monoxide molecule. Therefore, if we have 4.5 moles of oxygenated hemoglobin (Hb(O2)4), all 4.5 moles of the species will release oxygen and bind to carbon monoxide. 
7 0
4 years ago
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