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astra-53 [7]
1 year ago
14

The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil

limeters? A) 1.15x 10-8 mm B) 1.15 1013 mm C) 1.15 10-7 mm D) 1.15 x 1017 mm
Chemistry
1 answer:
jasenka [17]1 year ago
5 0

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Given data

Average distance between nitrogen and oxygen atoms: 115 pm

Step 2: Convert the distance to meters (SI base unit)

We will use the conversion factor 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Convert the distance to millimeters

We will use the conversion factor 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

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3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
2 years ago
Calculate the mass, in grams, of 2.74 l of co gas measured at 33°c and 945 mmhg.
nalin [4]
<span>T = 33 + 273 =306 K moles CO = pV/RT = 1.24 x 2.74 / 0.08206 x 306=0.135 mass CO = 0.135 mol x 28.01 g/mol=3.78 g</span>
4 0
1 year ago
Predict your observation if a magnesium ribbon is put in potassium hydroxide solution and tested with a lighted wooden splinter.
koban [17]

mg reacts with the acid and produce visible bubbles of hydrogen gas. they will produce audible pops as the hydrogen ignites. ... Hydrochloric acid is a corrosive liquid. Hydrogen gas is explosive.

5 0
2 years ago
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

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

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
Read 2 more answers
A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne
Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0
2 years ago
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