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loris [4]
3 years ago
14

(i need help please ill give brainliest if u get it right)* the questions and everything are in the picture below*

Chemistry
1 answer:
NemiM [27]3 years ago
7 0

Explanation:im not really sure about this one but i think it is either c or b

ANSWER:C im pretty sure.sorry if wrong

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4. Why can't the subscripts be changed in a chemical equation in chemistry
neonofarm [45]
If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is
4 0
3 years ago
The conjugate acid of ch3nh2 is ________. the conjugate acid of ch3nh2 is ________. ch3nh2+ ch3nh2 ch3nh3+ ch3nh+ none of the ab
Verizon [17]
The conjugate acid of ch3nh2 is ch3nh3+<span>.
</span>For example methylamine in water chemical reaction: 
CH₃NH₂(aq)+ H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq).
According to  Bronsted-Lowry theory acid are donor of protons and bases are acceptors of protons (the hydrogen cation or H⁺). Methylamine (CH₃NH₂) is Bronsted base and it can accept proton and become conjugate acid (CH₃NH₃⁺).
3 0
3 years ago
If the pH is 3.94 what is the concentration of H
solong [7]

Answer:

If pOH is 3.05, pH is 10.95 or

[H+]=1.122⋅10−11 mol/L

Explanation:

pH=−log(H+)Now, what is H+

if your pH is 10.95 (since pH+pOH=14, your pH=10.95).

(H+)=10 −10.95

(H+)=1.122⋅10 −11

This is your hydrogen ion concentration: [H+]=0.00000000001122  mol/L

7 0
3 years ago
How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water
Zarrin [17]

Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

<u>Find the moles of each reactive</u>

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

<u>Divide by the coefficient</u>

<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

The smallest values are for H_2, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

6 0
3 years ago
What is the concentration of the base (KOH) in the following titration, given the data below? 37.0 M 16.95 M 0.75 M 0.45 M
Bingel [31]
Your answer should be 0.75
8 0
3 years ago
Read 2 more answers
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