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Ad libitum [116K]
3 years ago
10

2.0 cm + 2.6 cm + 4.2 mm = Please show work

Chemistry
1 answer:
Setler [38]3 years ago
3 0
Hi ;)!
So:
4.2 mm (/10) = 0.42 cm

2.0 cm + 2.6 cm + 0.42 cm =  5.02 cm ;)
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4 years ago
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A system is composed of 7.14 grams of Ne gas at 298 K and 1 atm. When 2025 joules of heat are added to the system at constant pr
dlinn [17]

Answer:

(a) 1 atm, 8.72 L and 298 K respectively.

(b) 1 atm, 16.72 L and 573.35 K respectively.

(c) \Delta U=1215J

Explanation:

Hello,

(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\  \\V=8.72L

Thus, initial conditions of pressure volume and temperature are 1 atm, 8.72 L and 298 K respectively.

(b) Since the process was carried out at constant pressure, the work is defined as:

W=P(V_2-V_1)

Thus, the final volume is:

V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L

And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:

Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K

Therefore, final volume is 16.72 L, final pressure is also 1 atm and final temperature is 573.35 K for this expansion process.

(c) Finally, the change in the internal energy is computed via the first law of thermodynamics:

Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J

Best regards.

8 0
3 years ago
Pls help!!! I will mark you as brainliest if you answer the question!!
SVEN [57.7K]

1 — Element D and A ( Which are Sodium and aluminium )

2 — ( 2 + 8 + 3 = 13 electrons total ) Element A Because it's atomic number is 13.

3 — Element E Is stable. ( Which is Argon )

( Note, Elements which has 8 election on its outermost cell is stable. ( Helium is exception which is a noble gas but have 2 electrons in outermost cell )

4 — Element F , which is hydrogen. ( Hydrogen is the only element to not have any neutron )

5 — Element D ( Which is sodium )

6 — The element F ( Which is hydrogen ) Don't contain any neutron.

________________________________

5 0
3 years ago
4. What do radio telescopes use to gather and focus radio waves?
Tom [10]
I think it would be C
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4 years ago
Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO 3 (s) PbO (s) CO 2 (g) ________ grams of lead (I
blsea [12.9K]

Answer:

We will have 7.30 grams lead(II) oxide

Explanation:

Step 1: Data given

Mass of lead (II)carbonate = 8.75 grams

Molar mass PbCO3 = 267.21 g/mol

Step 2: The balanced equation

PbCO3 (s) ⇆ PbO(s) + CO2(g)

Step 3: Calculate moles PbCO3

Moles PbCO3 = mass / molar mass

Moles PbCO3 = 8.75 grams / 267.21 g/mol

Moles PbCO3 = 0.0327 moles

Step 4: Calculate moles PbO

For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2

For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO

Step 5: Calculate mass PbO

Mass PbO = moles PbO * molar mass PbO

Mass PbO = 0.0327 moles * 223.2 g/mol

Mass PbO = 7.30 grams

We will have 7.30 grams lead(II) oxide

7 0
4 years ago
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