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3 years ago

13

Please HELP, the question is attached.

1 answer:

Nezavi [6.7K]3 years ago

7 0

Try this solution, all the details are described in the attached picture.

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How much heat is absorbed during production of 112 g of NO by the combination of nitrogen and oxygen?

djverab [1.8K]

From the given chemical equation we see that 43 kcal of energy is needed for every 2 moles of NO. First let us calculate the moles of NO with a molar mass of 30 g/mol.

moles NO = 112 g / (30 g/mol) = 3.73 mol

So the total heat absorbed is:

heat = (43 kcal / 2 mol) * 3.73 mol

<span>heat = 80.195 kcal</span>

7 0

3 years ago

Given the reaction: Mg(s) + 2 AgNO3(aq) → Mg(NO3)2(aq) + 2 Ag(s) Which type of reaction is represented?

Papessa [141]

Mg(s) + 2 AgNO₃(aq) = Mg(NO₃)₂(aq) + 2 Ag(s)

reaction is :<span> single replacement

hope this helps!</span>

6 0

4 years ago

You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam

katovenus [111]

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression $d=\frac{m}{V}$ where m is the mass and V is the volume, therefore:

- Density of the cube A:

$d_{A}=\frac{m_{A}}{V_{A}}$

- Density of the cube B:

$d_{B}=\frac{m_{B}}{V_{B}}$

- Density of the cube C:

$d_{C}=\frac{m_{C}}{V_{C}}$

Solving for mass:

$m_{A}=d_{A}*V_{A}$

$m_{B}=d_{B}*V_{B}$

$m_{C}=d_{C}*V_{C}$

And all the three cubes have the same mass, so:

$m_{A}=m_{B}=m_{C}$

Therefore:

$d_{A}*V_{A}=d_{B}*V_{B}$ (Eq.1)

$d_{A}*V_{A}=d_{C}*V_{C}$ (Eq.2)

Solving for $d_{1}$ in Eq.1:

$d_{A}=d_{B}\frac{V_{B}}{V_{A}}$

Replacing values for the volume:

$d_{A}=d_{B}\frac{16.7mL}{25.9mL}$

$d_{A}=d_{B}*0.64$

As we know the density of the aluminum is $2.7\frac{g}{cm^{3}}$ , so replacing this value for $d_{B}$ :

$d_{A}=2.7\frac{g}{mL}*0.64$

$d_{A}=1.728\frac{g}{mL}$

that is the density of the magnesium.

Solving for $d_{C}$ in Eq.2:

$d_{C}=d_{A}\frac{V_{A}}{V_{C}}$

$d_{C}=d_{A}\frac{25.9mL}{4.29mL}$

$d_{C}=d_{A}*6.04$

$d_{C}=1.728\frac{g}{mL}*6.04$

$d_{C}=10.4\frac{g}{mL}$

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

8 0

4 years ago

Can someone help me here?

tiny-mole [99]

The cell model/theory is a big one

the atomic model/theory

4 0

4 years ago

What is 21.97 to three significant figures

antiseptic1488 [7]

21.97

Sig Figs

4

Decimals

2

Scientific Notation

2.197 × 101

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers