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4 years ago

Please answer the following on the pictureASAP PLZZZ

Chemistry

1 answer:

maxonik [38]4 years ago

3 0

30. Atomic Number is 17

31. Chlorine

32. 35.45

33. Nonmetal

34. 3 energy levels

35. 7 Valence electrons

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The volume of a fixed amount of gas is doubled, and the absolute temperature is doubled. According to the ideal gas law, how has

neonofarm [45]

Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

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3 years ago

What is the central conflict in "Ellen foster"?

photoshop1234 [79]

major conflict. Ellen continually suffers abuse by her neglectful caretakers and searches for a stable home and loving family. rising action. Ellen is placed in a number of temporary homes, all of which are unhappy, and she longs for a home where she is loved and cared for.

7 0

3 years ago

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the alloy is 2.324 J/g°C

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3 years ago

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2 AlCl 3 ( aq )

Dovator [93]

Answer:

The volume of hydrogen gas produced at STP is 4.90 liters.

Explanation:

$2Al (s) + 6 HCl (aq)\rightarrow 2 AlCl_3 (aq) + 3 H_2 (g)$

Moles of aluminium = $\frac{3.60 g}{27 g/mol}=0.1333 mol$

According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.

Then 0.1333 moles of aluminium will give:

$\frac{3}{2}\times 0.1333 mol=0.2 moles$ of hydrogen gas

Volume of 0.2 moles of hydrogen gas at STP = V

Temperature at STP = T = 298.15 K

Pressure at STP = P = 1 atm

n = 0.2 mol

PV = nRT (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.2mol\times 0.0821 atm L/mol K\times 298.15 K}{1 atm}=4.8956 L\approx 4.90 L$

The volume of hydrogen gas produced at STP is 4.90 liters.

4 0

3 years ago

An object weighting 9.6 grams is placed in a graduated cylinder displacing the volume from 10.0mL to 13.2 mL

grigory [225]

The volume of object is 3.2 ml

<h3>Explanation:</h3>

Given:

Mass of the object = M = 9.6 g

Initial volume of liquid: $V_1 = 10.0\ ml$

Final volume of liquid after displacement: $V_2 = 13.2\ ml$

Total volume of the displaced object inside a graduating cylinder will be given as difference between the final volume and initial volume of the expanding object.

$V = V_2-V_1 = 13.2 -1 0.0 = 3.2\ ml$

V = 3.2 ml

3 0

3 years ago