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Stels [109]
3 years ago
10

1. Express in conventional notation (no exponents) in the space provided within the

Chemistry
1 answer:
mina [271]3 years ago
7 0

Answer:

a) 320: two significant figures.

b) 2,366: four significant figures.

c) 73.0: three significant figures.

d. 532.5: four significant figures.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:

a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.

b) 2,366: four significant figures.

c) 73.0: three significant figures, because the zero is followed by the decimal place.

d. 532.5: four significant figures.

Regards!

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PLEASE HELP!!
grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

3 0
3 years ago
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0
3 years ago
180m/173 Tl -> 180/173 Tl + ? Express your answer as a nuclear equation ​
Law Incorporation [45]

Answer: The nuclear equation is ^{180}_{173}Tl \rightarrow ^{180}_{173}Tl + ^{0}_{0}\gamma.

Explanation:

A nuclear reaction in which a heavy particle splits into another particle along with release of energy is called a nuclear fission reaction.

For example, ^{180}_{173}Tl \rightarrow ^{180}_{173}Tl + ^{0}_{0}\gamma

Here, energy is radiated in the form of gamma radiation.

Thus, we can conclude that the nuclear equation is ^{180}_{173}Tl \rightarrow ^{180}_{173}Tl + ^{0}_{0}\gamma.

4 0
2 years ago
What's the equation for the reaction of Trifluoro bromide with Ammonia gas. ( brainliest assured )​
sp2606 [1]

Answer:

Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.

Explanation:

please mark me as brainliest thank you

5 0
2 years ago
Which one of the following is not a property of a base?
AlekseyPX

Answer:

option B is correct

Explanation:

5 0
3 years ago
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