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saw5 [17]
3 years ago
10

Your observation and assessment data should help you _______ your curriculum A. change B. modify and individualize C. evaluate t

he effectiveness of D. complement
Chemistry
2 answers:
Degger [83]3 years ago
8 0
The answer is b. Modify and individualize
joja [24]3 years ago
4 0
Hello,
I believe the answer is C. A and B say that it would help to change it but when we think about it data and opinions are not what people use to change a curriculum or program. Mainly they use surveys, which they didn't use here. Option D doesn't make any sense, data doesn't complement a program it supports some sort of theory which leads us to C. C is the only option where they use the data for effectiveness. Hope this helps!
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The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2).
Travka [436]
2 NI₃= N₂ + 3 I₂

2 x 394.71 g --------------- 3 x 253.80 g
3.58 g ---------------------- ( mass  of I₂ )

3.58 x 3 x 253.80 / 2 x 394.71 =

2725.812 / 789.42 => 3.4529 g of I₂

1 mole I₂ --------------- 253.80 g
?? ----------------------- 3.4529 g

3.4529 x 1 / 253.80 => 0.0136 moles of I₂

Answer C

hope this helps!
8 0
3 years ago
Read 2 more answers
2 Na2O + CO = Na2CO3 + Na
Vikki [24]

Moles of Na_2O

  • Given mass/Molar mass
  • 35/62
  • 0.56mol

2 mols Na_2O produces 1 mol sodium carbonate.

1 mol sodium oxide produces 0.5mol sodium carbonate

Moles of sodium carbonate

  • 0.56(0.5)
  • 0.28mol

Molar mass

  • 2(23)+12+3(16)
  • 46+12+48
  • 106g/mol

Mass

  • 0.28(106)
  • 29.7g
8 0
2 years ago
Which of the following would you predict to have a positive value for º ?
alexdok [17]
Given:

AgCl (s) ===> Ag+ (aq) + Cl- (aq)
- negative entropy

H2O(g) ===> H2O(l)
- positive entropy

2Na (s) + Cl2 (g) ===> 2NaCl (s)
- positive entropy

Br2(l) ===> Br2 (s)
-positive entropy

They are identified to have positive or negative values of entropy based on the phases of the reactants and products. 
8 0
4 years ago
What is produced during the replacement reaction of ba(no3)2 and na2so4? 2bana 2no3so4 2nano3 baso4 nano3 baso4 bana2 (no3)2so4
Cloud [144]

2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Procedures involved:

The cations or anions may transfer positions in this twofold replacement/displacement reaction, which results in AB + CD AD + CB. In such a reaction, water, an insoluble gas, or an insoluble solid must be one of the byproducts (precipitate). The reaction in question has the following molecular equation:

2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Double displacement:

When two atoms or groups of atoms swap positions, a double displacement reaction occurs, creating new compounds. Typically, aqueous solutions are where it happens.

Na2SO4 + BaCl2 BaSO4 + 2NaCl is an example of a double displacement reaction.

To learn more about double displacement refer :brainly.com/question/23918356

#SPJ4

8 0
2 years ago
A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go
4vir4ik [10]

Answer :  The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of NH_4Cl = 26.5 g

Mass of NaOH = 10 g

Molar mass of NH_4Cl = 53.5 g/mole

Molar mass of NaOH = 40 g/mole

First we have to calculate the moles of NH_4Cl and NaOH.

\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

NH_4Cl+NaOH\rightarrow NH_4OH+NaCl

From the balanced reaction we conclude that

As, 1 mole of NaOH react with 1 mole of NH_4Cl

So, 0.25 moles of NaOH react with 0.25 moles of NH_4Cl

From this we conclude that, NH_4Cl is an excess reagent because the given moles are greater than the required moles and NaOH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of NH_4Cl.

\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl

\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g

Therefore, the amount of reactant left in excess is 13.1075 grams.

5 0
3 years ago
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