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yuradex [85]
3 years ago
6

What heats the mantle?

Chemistry
1 answer:
storchak [24]3 years ago
7 0

Answer:

Fire

Explanation:

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How many moles of calcium atoms are in 77.4g of Ca?
lina2011 [118]

Answer:

1.93 mol Ca

Explanation:

Use Calcium's molar mass to convert g of Ca into mol of Ca.

6 0
2 years ago
The lithospheric plates float on top of the _______, which is a weak part of the mantle that flows slowly.
alexandr1967 [171]
<span>The lithospheric plates float on top of the _______, which is a weak part of the mantle that flows slowly.
</span>
asthenosphere
7 0
3 years ago
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane i
Kipish [7]

Answer:

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

8 0
3 years ago
V7
dangina [55]

Answer:

sodium chloride

Explanation:

3 0
2 years ago
1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?
RSB [31]

Answer : The number of molecules present in nitrogen gas are, 3.48\times 10^{13}

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of N_2 gas = 1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm      (1 atm = 760 mmHg)

V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 0.0^oC=273+0.0=273K

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

(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

Therefore, the number of molecules present in nitrogen gas are, 3.48\times 10^{13}

8 0
3 years ago
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