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Travka [436]
3 years ago
8

33) How many significant figures does the number 40230 have? *

Chemistry
1 answer:
Lyrx [107]3 years ago
4 0

Answer:

5

Explanation:

they are all significant All non-zero numbers ARE significant

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You have a 1.8 M solution of HCI. You need to make a 0.70 M solution with a volume of 550 mL. How many mL of the stock solution
Arte-miy333 [17]

The volume of the stock solution needed is 213.88 mL to get new concentration.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Determination of the volume of stock solution.

Volume of diluted solution (V₂) =550 mL

Molarity of diluted solution (M₂) =0.70 M

Molarity of stock solution (M₁) = 1.8 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

1.8 M  × V₁ = 0.70 M × 550 mL

V₁ = 213.88 mL

Thus, the volume of the stock solution needed is 213.88 mL.

Learn more about the molarity here:

brainly.com/question/2817451

#SPJ1

5 0
2 years ago
How long will a current of 0.995 A need to be passed through water (containing H2SO4) for 5.00 L of O2 to be produced at STP
DIA [1.3K]

Answer:

24 hours

Explanation:

The computation is shown below:

The needed mole of O_2 is

= 5 ÷22.4 = n

Also 1 mole of O_2 required four electric charge

Now the charge needed is

= n × 4 × 96,500 C

= 4 × 96,500 × 5 c ÷ 22.4

= 86160.714 C

Now

q = i t

t = q ÷ i

= 86160.714 C ÷ 0.995

= 86593.7 seconds

= 24 hours

Hence, the correct option is A.

3 0
3 years ago
How many molecules of sodium chloride are in 2.5 moles??
sergiy2304 [10]

Answer:

1.5x1024

Explanation:

7 0
2 years ago
A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the t
Mariulka [41]

Answer:

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

<em> Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>

Explanation:

Using the formula of heat, Q = mc∆T  

where Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron  = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C  = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

5 0
2 years ago
The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However
Murljashka [212]

Answer:

Reverse the 2 NO_2 \longrightarrow 2 NO + O_2  reaction

Explanation:

Reactions:

2 NO_2 \longrightarrow 2 NO + O_2

N_2O_4 \longrightarrow 2 NO + O_2

Overall:

N_2O_4 \longrightarrow 2 NO_2

As can be seen, in the overall reaction we have N_2O_4 in the reactants like in the second reaction and NO_2 in the products. The NO_2 is in the first reaction but as a reactant so we need to reverse that reaction:

2 NO + O_2 \longrightarrow 2 NO_2

N_2O_4 \longrightarrow 2 NO + O_2

Combining:

N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2

N_2O_4 \longrightarrow 2 NO_2

4 0
3 years ago
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