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Sergio039 [100]

3 years ago

15

Which represents a balanced chemical equation?

2 answers:

Serhud [2]3 years ago

8 0

Answer:

C

Explanation:

Reactant part has 2 Mg and 2 O, product part has 2 Mg and 2 O

romanna [79]3 years ago

4 0

I think it’s b but i could be wrong

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Which sample of matter is classified as a solution? 1. H2O(s) 2. H2O(l) 3. CO2(g) 4. CO2(aq)

lys-0071 [83]

Answer : Option 4) $CO_{2}_{(aq)}$

Explanation : $CO_{2}_{(aq)}$ is the only sample of matter which can be classified as a solution. As the solution can be defined as a liquid mixture which contains a minor component (the solute) that is uniformly distributed within the major component (the solvent).

In this case, the solute is $CO_{2}$ which is dissolved in water which acts as an solvent. Also, it has a subscript which is aq. which means aqueous, is often given to the solution in which the solvent is water.

Therefore, $CO_{2}_{(aq)}$ is the correct answer.

4 0

3 years ago

Read 2 more answers

Juan observes a material in the rocks of a hilluan hammers off a piece and then examines the pieces with a hana tena He

Nata [24]

Answer:

I definitely think he mostly observed that it was clear in color.

Explanation:

5 0

3 years ago

Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.

Serggg [28]

Answer:

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

Explanation:

$Q=mc\Delta T$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have mass of copper = m = 25.3 g

Specific heat of copper = c = 0.385 J/g°C

ΔT = 39°C - 22°C = 17°C

Heat absorbed by the copper :

$Q=25.3 g\times 0.385 J/g^oC\times 17^oC=165.59 J$

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

5 0

3 years ago

In the compound CO2, how many lone pairs are on the central atom?

Cloud [144]

There are 4 lone pairs of electrons present in the carbon dioxide molecule

4 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$q=m\times c\times \Delta T$

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

5 0

3 years ago