How many moles of hydrogen are there in one mole of h2o?

How many molecules of nitrogen monoxide are in a 22.5 gram sample?

poizon [28]

Okay, so you need to start by finding the molar mass (grams in one mole) of nitrogen monoxide (NO).
N=14g
O=16g
So we know that NO has a molar mass of 30 grams, then just divide your given mass of 22.5 grams by the molar mass of 30 grams to find the number of molecules in your sample. The answer should be .75 molecules. Hope this helps!

3 0

3 years ago

What will happen if we put iron nails with sodium bicarbonate in a closed flask?

rjkz [21]

Answer:

You will be able to watch the slow dissolution of the nail (probably a steel nail and not iron) and the change in colour of the solution.

Explanation:

I believe not 100% sure.

5 0

2 years ago

A sample of a compound contains 60.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula?

Inessa [10]

A sample of a compound contains 60.0 g C and 5.05 g H.

divide by molar mass of C(12) and H(1) to get molar ratio

C: 60/12=5 and H: 5/1=5

so C:H=5:5=1:1

total molar mass=78

divide by 1C+1H to find the formula: 78/(12+1)=78/13=6

compound is C6H6

5 0

4 years ago

Read 2 more answers

Which of the following statements is true about the specific heat capacity of a substance?

Ludmilka [50]

Answer:

Option A

Explanation:

An intensive property is a bulk property, meaning that it is a local physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, T; refractive index, n; density, ρ; and hardness of an object,specific heat, η.

Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, specific heat and many others.

6 0

3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago