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3 years ago

How many grams of NaNO3 can be dissolved in 200mL of water at 40 degrees Celsius?

Chemistry

1 answer:

Tju [1.3M]3 years ago

6 0

Answer:

I think the answer is, 100.

Explanation:

but i dont know try

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What is the theoretical yield of sodium oxide (Na₂O) in grams when 20.0g of calcium oxide (CaO) reacts with an excess amount of

ad-work [718]

Answer:

22.1g

Explanation:

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

CaO + 2NaCl —> Na2O + CaCl2

Next, we shall determine the mass of CaO that reacted and the mass of Na2O produced from the balanced equation.

This is illustrated below:

Molar mass of CaO = 40 + 16 = 56g/mol

Mass of CaO from the balanced equation = 1 x 56 = 56g

Molar mass of Na2O = (23x2) + 16 = 62g/mol

Mass of Na2O from the balanced equation = 1 x 62 = 62g

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Finally, we can determine the theoretical yield of Na2O as follow:

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Therefore, 20g of CaO will react to produce = (20 x 62)/56 = 22.1g of Na2O.

Therefore, the theoretical yield of Na2O is 22.1g

6 0

3 years ago

Read 2 more answers

True/False: When a substance is heated, the change in energy causes the bonds between particles to become stronger.

Rzqust [24]

Answer:

false

Explanation:

the energy causes the bonds to become looser

5 0

4 years ago

Read 2 more answers

the amount of solar energy reaching 1 m² of earths surface each second is 1.366 x 10^3 joules. if this energy was converted into

Advocard [28]

I think is 6.588579795x10^13 Kg because the equation is E=mc^2 and E is your Joules and c^2= 9x10^16 so m=(c^2)/E

4 0

4 years ago

What is the [H+] in a solution with pOH of 0.253?

sergey [27]

PH + pOH = 14

pH + 0.253 = 14

pH = 14 - 0.253

pH = 13.747

[ H+] = 10 ^ -pH

[ H+ ] = 10 ^- 13.747

[ H+ ] = 1.790x10⁻¹⁴ M

hope this helps!

4 0

3 years ago

Read 2 more answers

The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K

8 0

3 years ago