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3 years ago

Calculate the number of formula units in 1.87 mol of NH4Cl

1 answer:

7 0

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What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?

Vladimir79 [104]

Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2 -------> 32 g O2 -----------> 36 g H2O
↓ ↓ ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.

4 0

3 years ago

A salt is made by adding an excess of an insoluble metal oxide to an acid.

Blizzard [7]

Answer:

Filtration

Explanation:

Filtration is used to separate solid substances from liquids or large molecules from small molecules. Since, Acid is a liquid and metal oxide is solid so Filtration can be used to separate these two.

3 0

3 years ago

True or False. Isotopes have different chemical properties. Explain why?

Maurinko [17]

Answer:

False

Explanation: isotopes generally have the same physical and chemical properties because they have the same numbers of protons and electrons.

4 0

3 years ago

The equations ? It’s for a chemistry class, and he said we need to explain and write down the math problems?

alina1380 [7]

Answer:

I think this is because math and chemistry go together and the math problems are science related.

Explanation:

3 0

2 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

4 years ago