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3 years ago

What are 2 interactions between the earth spheres and the carbon cycle?

Chemistry

1 answer:

Fed [463]3 years ago

6 0

Answer:

Explanation:

The spheres interact with each other, and a change in one area can cause a change in another. Humans (biosphere) use farm machinery manufactured from geosphere materials to plow the fields, and the atmosphere brings precipitation (hydrosphere) to water the plants. The biosphere contains all the planet's living things.

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Describe what golgi looks like.

Margaret [11]

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2 years ago

given the following reaction: CuO (s) + H2 (g) to Cu(s) + H2O (g). If 250 L of hydrogen gas are used to reduce copper (II) oxide

QveST [7]

24g is needed no explanation

4 0

2 years ago

What is the name of the process plants use to make the energy (food) they need?

scZoUnD [109]

The answer you looking for is D

8 0

3 years ago

Read 2 more answers

Using the balanced equation below, how many moles of water can be produced when 4 grams of Hydrogen react? 2H2(g)+O2(g) —> 2H

gavmur [86]

Answer:

4 × 10 g

Explanation:

Step 1: Write the balanced equation

2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)

Step 2: Calculate the moles corresponding to 4 g of H₂

The molar mass of H₂ is 2.02 g/mol.

4 g × 1 mol/2.02 g = 2 mol

Step 3: Calculate the moles of H₂O produced from 2 moles of H₂

The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.

Step 4: Calculate the mass corresponding to 2 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

2 mol × 18.02 g/mol = 4 × 10 g

7 0

2 years ago

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

2 years ago