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xenn [34]
3 years ago
14

How to make baking powder

Chemistry
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

To make baking powder, combine half a teaspoon of cream of tartar and a quarter teaspoon of bicarbonate of soda.

hope it helps;)

deff fn [24]3 years ago
4 0

Answer:

with baking soda

Explanation:

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What is the m of a solution where 0.500 moles of a salt are dissolved in 100.0 ml of solution? 25.0m 5.00m 50.0m o.500m 2.50m?
My name is Ann [436]
Molarity=moles/litre
molarity=0.5/0.1
molarity=5.00m
3 0
3 years ago
Is color change a physical or chemical change?
bagirrra123 [75]
It's a chemical change because it's changing the object's property. 
3 0
3 years ago
1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en
Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles      

n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles

m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

\% = \frac{R_{r}}{R_{T}}*100

<u>Donde</u>:

R_{r}: es el rendimiento real

R_{T}: es el rendimiento teórico

\% = \frac{3,5}{5,043}*100 = 69,4

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

5 0
3 years ago
The combustion reaction of magnesium results in what compound?
Tasya [4]
C. MgO because mg+o= MgO
7 0
3 years ago
Read 2 more answers
How many grams of H2O are needed to produce 150 g of Mg(OH)2? (Molar mass: H2O = 18.02 g/mol; Mg(OH)2 = 58.33 g/mol )
Nuetrik [128]

Answer:

92.6

Explanation:

6 mol x 18.02 g of H2o --> 3 mol x 58.33 g  Mg(OH)2

108.12 g of h2o --> 174.99 of Mg(OH)2

g of H2O is 150 g of Mg(OH)2

150g x 108.12g / 174.99 =

92.67

7 0
3 years ago
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