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Anton [14]
2 years ago
12

¿Cuántos moles hay en 350 g de NaNO3?

Chemistry
1 answer:
olga55 [171]2 years ago
4 0

Answer:

3 moles

Explanation:

So, approximately 3 moles of NaNO3 can be obtained, by reacting 253 grams of Na2CrO4. Also, the number of moles is a dimensionless quantity.

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Please help me with this question!
asambeis [7]
I cant help u sorry .......
3 0
3 years ago
Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
ivanzaharov [21]
<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol

</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
    reactants                         products

products- reactants:</span><span>

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
<span>
</span>






6 0
2 years ago
3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?
Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J

Thus, the total energy turns out to be:

Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J

Best regards!

5 0
2 years ago
What is the volume of 0.1 mole of methane (CH4) ? (One mole of any gas occupies 22.4 L under certain conditions of temperature a
Veronika [31]

Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.

For 0.1 mole of methane.

0.1/1 × 22.4 = 2.24 L

0.1 mole of methane occupy 2.24 L volume.

8 0
3 years ago
Read 2 more answers
How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
Flauer [41]
I’m assuming you mean barium nitrite, Ba(NO2)2.

First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.

45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
4 0
2 years ago
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