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aniked [119]
1 year ago
5

Perform the following opeartiona dn express the answer in scientific notation 3.14x10^-2/2.65x10^-7

Chemistry
1 answer:
Firdavs [7]1 year ago
4 0

Answer:

1.18 x 10⁵

Explanation:

3.14 x 10⁻² / 2.65 x 10⁻⁷ = 1.18 x 10⁵

When performing division, the final answer should have the same number of significant figures as the number with the smallest number of sig figs. In this case, both of the given numbers have 3 sig figs, meaning the final answer should have this many as well.

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If 57.3 l of 0.497 m koh is required to completely neutralize 39.5 l of a CH3COOH solution. What is the molarity of the acetic a
bearhunter [10]

The molarity of the solution will be 0.72 m.

The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.

The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.

Given data:

V_{1} =57.3 L\\V_{2} = 39.5 L\\M_{1} = 0.497 m\\\\M_{2} = ?

Molarity can be determined by the formula:

M_{1} V_{1} = M_{2} V_{2}

where, M is molarity and V is volume.

Put the value of given data in above equation.

57.3 × 0.497 m = M × 39.5 L

M = 0.72 m

Therefore, the molarity of the solution will be 0.72 m

To know more about molarity

brainly.com/question/18648803

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6 0
2 years ago
If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?
GenaCL600 [577]
K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K          +            10KNO₃  ⇒  6K₂O + N₂
2mol        :            10mol
3,97mol   :           1,21mol
                             limiting reagent

KNO₃ is limiting reagent

5 0
3 years ago
If a match were placed in the cone close to the barrel of the gas burner will it ignite
Pani-rosa [81]
I think it would seeing as you typicaly have to use a match to light a gas burner
8 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
Freezing point<br> Where would you place this property in the Venn diagram?
Marat540 [252]
For the Venn diagram, freezing water should be in the physical category (A)
7 0
3 years ago
Read 2 more answers
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