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nydimaria [60]
2 years ago
12

Please help fast thank yuo

Chemistry
2 answers:
Masteriza [31]2 years ago
7 0

Answer:

what is your question tell may i can help you

dmitriy555 [2]2 years ago
6 0

Answer:

decrease as we go down a group I know ionisation energy increases from left to right across the periodic table

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When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h
Pavel [41]
NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃                   <span>+                        3O</span>₂                      ⇒<span>          2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol                     :                        3mol                   :             2mol
0.133mol             :                        0.116mol           :             0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
6 0
3 years ago
Three students are labeling organisms in their diorama. Zola wants to label the algae and mushrooms as photosynthetic. Saffi wan
Nookie1986 [14]

Answer:

Saffi only

Explanation:

I just took the test and that was the correct answer :)

3 0
2 years ago
Serine has pka1 = 2.21 and pka2 = 9.15. use the henderson-hasselbalch equation to calculate the ratio neutral form/protonated fo
malfutka [58]
Calculate the ratio by using Henderson-Hasselbalch equation:

pH = pKa + log [neutral form] / Protonated form

3.05 = 2.21 + log [neutral form] / [Protonated form]

3.05 - 2.21 = log [neutral form] / [Protonated form]

0.84 = log [neutral form] / [Protonated form]

[neutral form] / [protonated form] = anti log 0.84 = 6.91
8 0
3 years ago
A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0
OleMash [197]

<u>Answer:</u> The mass of original oxalic acid sample is 6.75 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_2O_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL

Putting values in above equation, we get:

2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0
3 years ago
Can you help me out plz and thank u​
Nonamiya [84]

Answer:

stirring

Explanation:

when you stir it spreads the item out more to be fully covered and dissolve faster. like putting sugar in tea, if you don' t stir it wont dissolve  fast

5 0
3 years ago
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