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2 years ago

What is the mole ratio of hydrogen to nitrogen in the following reaction: N2(g) + 3H2(g) →

Chemistry

1 answer:

Nuetrik [128]2 years ago

5 0

Answer:

3 mol H2/2 mol N2

Explanation:

Mole ratios are basically from the numbers/coefficients in front of the element

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What is the mass of a magnesium block that measures 2.00 cm

skad [1K]

Answer:

m = 31.284 grams

Explanation:

Given that,

The dimension of a magnesium block is 2.00 cm x 3.00 cm x 3.00 cm.

The density of magnesium is, d = 1.738 g/cm³

We need to find the mass of the magnesium block. We know that the density of an object is given by its mass per unit its volume. So,

$d=\dfrac{m}{V}\\\\\text{Where m=mass}\\\\m=d\times V\\\\m=1.738\ g/cm^3\times (2\times 3\times 3)\ cm^3\\\\m=31.284\ \text{grams}$

So, the mass of the block is 31.284 grams.

5 0

3 years ago

Where does energy in an ecosystem come from?

Novay_Z [31]

Answer:

Energy in an ecosystem comes from the Sun.

Explanation:

It is a major source of energy for organisms, plants and the whole ecosystem basically.

4 0

3 years ago

Read 2 more answers

11. Power can be defined as (2 points)

Ierofanga [76]

Answer:

the energy required to do work

4 0

3 years ago

How much mass defect is required to release 5.5 x 1020 J of energy?

Montano1993 [528]

Answer : The mass defect required to release energy is 6111.111 kg

Explanation :

To calculate the mass defect for given energy released, we use Einstein's equation:

$E=\Delta mc^2$

E = Energy released = $5.5\times 10^{20}J$

$\Delta m$ = mass change = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in above equation, we get:

$5.5\times 10^{20}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2$

$\Delta m=6111.111kg$

Therefore, the mass defect required to release energy is 6111.111 kg

5 0

3 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago