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3 years ago

Will mark brainliest!!

Chemistry

1 answer:

just olya [345]3 years ago

5 0

Answer:

magnesium= +2

aluminum= +3

phosphorous= -3

lithium= +1

fluorine= -1

Explanation:

As fluorine having seven valance electrons in its outer most shell with atomic number nine. And for completing its outer most shell it needs one more electron that is why it form ion of -1 value.

Now Aluminum has three electrons in its outer most shell with atomic number 13. So it need to lose these three electrons in order to be in stable state that is why it have ion in +3 state.

Same rule apply for other elements too.

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Can you calculate the energy needed to increase the temperature of 100kg of iron by 40°C?

VLD [36.1K]

Answer: 177600 J or 177.6kJ

Explanation:

For this problem, we need to use q = mcAT

we have m = 10000, and AT= 40. The specific heat of iron is 0.444.

So now we can plug it in: q = 10000*.444*40 = 177600 J or 177.6kJ

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3 years ago

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The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

Answer: The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

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3 years ago

A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter yo

ra1l [238]

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

$6.35ft \times \frac{1m}{3.28ft} = 1.94m$

Step 3: Calculate the volume of water (V)

We will use the following expression.

$V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}$

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

$4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL$

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3 years ago

Three different sub-atomic particles are described below: Particle X: Uncharged Particle Y: Carries positive charge Particle Z:

bazaltina [42]

The statement which is true about the three sub atomic particles is

X is a neutron,Y is a protons and z is an electron

Explanation

Atom has three sub atomic particles,that is neutrons,protons and electrons.

Neutrons have no electric charge,protons are positively charged while electrons are negatively charged.
Both neutrons and proton are located in the nucleus while electrons are located outside the nucleus.
since electrons are located outside the nucleus they orbit around the nucleus.

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3 years ago

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Which equation shows conservation of atoms?

makvit [3.9K]

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3 years ago

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