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Travka [436]
2 years ago
7

can someone help me with these? I just started chemistry this semester, and since it's involving math I don't see myself doing g

ood.

Chemistry
1 answer:
V125BC [204]2 years ago
4 0

Any non zero digits are significant. (120 has 2 sig figs)

Zeroes between significant digits are significant. (103 has 3 sig figs)

Zeroes only count if there is a decimal point and if they are following a significant digit. (.600 has 3 sig figs; .067 has 2 sig figs)

1) <u>1278.50</u> = 6 sig figs

2) <u>12</u>0000 = 2 sig figs

3) <u>90027.00</u> = 7 sig figs

4) 0.00<u>53567</u> = 5 sig figs

5) <u>67</u>0 = 2 sig figs

6) 0.00<u>730</u> = 3 sig figs

That's all I'm gonna do so yeah.

Now for the rounding, it's basically rounding but it must have the x amount of sig figs after rounding. It's pretty hard to explain.

19, I'm not really sure. Might be 120 with a decimal point at the end.

20) 5.457, alright so here, since we are rounding to three significant digits, we look at the third sig fig. The third sig fig here would be 5. Then we just round. Since the number next to 5 is 7, we round 5 up to 6.

So 5.457 rounded to 3 sig figs would be 5.46

21) 0.0008769, the third sig fig here would be 6. You round 6 up to 7 because the number next to it is 9.

So 0.0008769 rounded to 3 sig figs would be 0.000877

So I hope you get it, I'm gonna let you do the rest.

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For dilute aqueous solutions in which the density of the solution is roughly equal to that of the pure solvent, the molarity of
EleoNora [17]

Explanation:

Given: <u>Molarity = 0.01 M</u>

Let the volume of the solution = 1 L

Hence, moles = 0.01 moles (Molarity*Volume)

Molar mass of urea = 60 g/mol

So, mass of urea = moles x molar mass = 0.01 moles x 60 g/mol = 0.6 g = 0.0006 kg ,

Given: Density of solvent = Density of solution = 1 Kg/liter (Water).

So,

The mass of solution = vol x density = 1 L x 1 kg/L= 1 kg

Also,

Mass of solution = Mass of solute + Mass of solvent

1 kg= 0.0006 kg + mass of solvent

Mass of solvent = 1-0.0006 = 0.9994  kg

Molality is the moles of solute present in 1 kg of the solvent. So,

<u>Molaity = ( 0.01/0.9994)  ≈ 0.01  m</u>

<u>Molarity = Molality</u>

Hence proved.

3 0
4 years ago
A balloon is filled with 12 L of air at a pressure of 2 atm. What is the volume of the balloon if the pressure is changed to 3 a
mihalych1998 [28]

Answer:

8L

Explanation:

Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant

P1V1= P2V2

P1 =  2atm, V1 = 12L ,

P2 = 3atm , V2 =

12 × 2 = V2 × 3

Divide both sides by 3

V2 = 24 ÷ 3

V2 = 8L

I hope this was helpful, please mark as brainliest

3 0
3 years ago
Be sure to answer all parts.
kondaur [170]

Answer:

154.44 Kg of H₂

Explanation:

We'll begin by converting 139 Kg of water to grams. This can be obtained as follow:

1 kg = 1000 g

Therefore,

139 kg = 139 kg × 1000 g / 1 Kg

139 Kg = 139000 g

Next, the balanced equation. This is illustrated below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the mass of H₂ that reacted and the mass of H₂O produced from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of H₂O = (2×1) + 16

= 2 + 16

= 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

4 g of H₂ reacted to produce 36 g of H₂O.

Next, we shall determine the mass of H₂ needed to produce 139 kg (i.e 139000 g) of H₂O. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted to produce 36 g of H₂O.

Therefore Xg of H₂ will react to produce 139000 g of H₂O i.e

Xg of H₂ = (4 × 139000)/36

Xg of H₂ = 154444 g

Finally, we shall convert 154444 g to kg this can be obtained as follow:

1000 g = 1 Kg

Therefore,

154444 g = 154444 g × 1 Kg / 1000 g

154444 g = 154.44 Kg

Thus, 154.44 Kg of H₂ is needed to produce 139 Kg of water H₂O

6 0
3 years ago
Need Help Please, ASAP:
Dafna11 [192]
Go to peak it helped me a lot
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4 years ago
Read 2 more answers
Which of these substances contributes to soap’s slippery texture?
kompoz [17]

Answer:

The fatty acids and glycerol contributes to soap’s slippery texture.

Explanation:

  • What we need to know is that every oil has a specific index for strength, cleaning, etc.
  • All of them are determined by the amount of certain fatty acid in the oil.
  • When we make soap, what we need to know is that oil has a special role to play in that soap.
  • So for example some of those oils are used to make bubble soap or creamy soap.
  • On the other side, there are some oils which are giving strength to soap.
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