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2 years ago

Balance the equation for the reaction in which sodium oxide reacts with water to form sodium hydroxide.

Chemistry

1 answer:

tensa zangetsu [6.8K]2 years ago

6 0

Answer:

$➢ \: Balance \: the \: equation \: for \: the \: reaction \\ in \: which \: sodium \: oxide \: reacts \\ with \: water \: to \: form \: sodium \: hydroxide.$

⇒ We have Na2O + H2O --> NaOH. We have 2 sodiums and 2 oxygens and 2 hydrogens on the left side, but only one of each on the right side.

Sodium Oxide + Water → Sodium Hydroxide

⇒ Na2O + H2O → 2NaOH .

Sodium oxide is used in ceramics and glasses. Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution.

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Is bleach liquid starch? Yes or No

Jobisdone [24]

Answer:

O it's not

Explanation:

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8 0

2 years ago

A 20.0 gram sample of an element contains 4.95 x 1023 atoms. what is the element?

Sonja [21]

You will want to find how many grams are in a whole mole so you know which element it is. To do this, find out how much of a mole you have.

4.95 x 10^23 atoms / 6.022 x 10^23 atoms (one whole mole of any element) = .8219860511 or ~82% of 1 mole

Now we know that, find what to multiply 20 g by to get the rest of the mole.

1 mole / .8219860511 mole = 1.216565657

20 g x 1.216565657 = ~24.33 g / mol

Now that you have grams per mole, you can look at the periodic table and the molar masses to see which this number is closely aligned.

Your answer is Magnesium (Mg), which has a molar mass of 24.305 g

7 0

3 years ago

A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?

Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a gas, therefore we can use the ideal gas law:

$PV=nRT$

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 $\frac{atm*L}{mol*K}$

n= ?

So, we can solve for "n" (moles):

$1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K$

$n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}$

$n=0.126~mol$

Now, we have to remember that the molar mass value has "g/mol" units. We already have the grams (9.33 g), so we have to divide by the moles:

$molar~mass=\frac{9.33~grams}{0.126~mol}$

$molar~mass=73.88\frac{grams}{mol}$

7 0

3 years ago

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af

Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0

3 years ago

Write a balanced equation for the following:

Ket [755]

(ANS1)— P4 + 5O2 ---> 2P2O5

(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20

(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4

7 0

3 years ago