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3 years ago

Can someone help Plz

Chemistry

1 answer:

Marrrta [24]3 years ago

6 0

Answer: A. 1.60 liters

Explanation:

$2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2$

As can be seen from the balanced chemical equation, 2 moles of $CO_2$ produce 1 mole of $O_2$ .

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus $2$ moles of $CO_2$ occupies $22.4\times 2=44.8L$ at STP and produce 1 mole of $O_2$ i.e. 22.4 L

44.8 L of $CO_2$ will produce 22.4 L of $O_2$

Thus $3.20L$ of $CO_2$ will produce $\frac{22.4}{44.8}\times 3.20=1.60L$

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If there were an element above fluorine in the periodic table what phase would you expect it to be in (at room temp)?

AysviL [449]

If there were an element above fluorine, its state would be a gas. This is because fluorine is located in the non-metal section of the periodic table which can all be found as a gas at room temperature.

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3 years ago

The elements in a row of the the periodic table

Aleks [24]

Answer:

Rows in the periodic table are called periods. As one moves from left to right in a given period, the chemical properties of the elements slowly change. Columns in the periodic table are called groups. Elements in a given group in the periodic table share many similar chemical and physical properties

Explanation:

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3 years ago

How many atoms are in 578g Na

borishaifa [10]

1.51 x 10²⁵atoms

Explanation:

Given parameters:

Mass of Na = 578g

Unknown:

Number of atoms = ?

Solution:

To find the number of atoms, we must first find the number of moles the given mass contains.

Number of moles = $\frac{mass}{molar mass}$

molar mass of Na = 23g

Number of moles = $\frac{578}{23}$ = 25.13moles

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25.13 mole of Na = 25.13 x 6.02 x 10²³atoms

This gives 1.51 x 10²⁵atoms of Na

Learn more:

Avogadro's constant brainly.com/question/2746374

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5 0

3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

3 years ago

Is this right????????????

zzz [600]

Answer: yea its right

Explanation:

5 0

3 years ago

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