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Anna007 [38]
2 years ago
10

Help Me Please Thank You

Chemistry
2 answers:
Maurinko [17]2 years ago
7 0
For quest or 4 it would be C. Im cant see all the answers on 5 so I’m not sure sorru
NISA [10]2 years ago
3 0
The Gulf Stream, warm current in the Atlantic
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If the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?
lana66690 [7]

Answer:

2, strong acid

Explanation:

Data obtained from the question. This includes:

[H+] = 0.01 M

pH =?

pH of a solution can be obtained by using the following formula:

pH = –Log [H+]

pH = –Log 0.01

pH = 2

The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.

Since the pH of the above solution is 2, it means the solution is a strong acid.

5 0
3 years ago
You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of
Virty [35]

Answer:

When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.

Aldolase cleaves a hexose into two trioses.

[See the image attached].

Asterisk indicates the label.

When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.

5 0
3 years ago
The solubility of calcium sulfate at 30°c is 0.209 g/100 ml solution. calculate its ksp.
inessss [21]
Answer is: Ksp for calcium sulfate is 2.36·10⁻⁴.
Balanced chemical reaction (dissociation):
CaSO₄(s) → Ba²⁺(aq) + SO₄²⁻(aq).
m(CaSO₄) = 0.209 g.
n(CaSO₄) = m(CaSO₄) ÷ M(CaSO₄).
n(CaSO₄) = 0.209 g ÷ 136.14 g/mol.
n(CaSO₄) = 0.00153 mol.
s(CaSO₄) = n(CaSO₄) ÷ V(CaSO₄).
s(CaSO₄) = 0.00153 mol ÷ 0.1 L = 0.0153 M.
Ksp = [Ca²⁺] · [SO₄²⁻].
[Ca²⁺] = [SO₄²⁻] = s(CaSO₄).
Ksp = (0.0153 M)² = 2.36·10⁻⁴.
8 0
2 years ago
HElPpp AsAPp I’ll mark you as brainlister!!
vichka [17]
A. There is no extra information
5 0
3 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
solong [7]

Answer:

89.55~\%~of~Fe_2O_3~in~the~sample

Explanation:

The first step in this reaction is the<u> converstion from Kg</u> of Fe <u>to</u>  <u>grams</u> of Fe_2O_3.

1.19x10^3~Kg~Fe~\frac{1000g~Fe}{1~Kg~Fe}~\frac{1~mol~Fe}{55.84~g~Fe}~\frac{1~mol~Fe_2O_3}{2~mol~Fe}~\frac{159.68~g~Fe_2O_3}{1~mol~Fe_2O_3}~\frac{1~Kg~Fe_2O_3}{1000~g~Fe_2O_3}

X~=~1.7x10^3~Kg~Fe_2O_3

Then we can calculate the <u>percentage</u> of  Fe_2O_3 in the sample:

\%~Fe_2O_3~=~\frac{1.7x10^3~Kg~Fe_2O_3}{1.9x10^3~Kg~Sample}*100

89.55~\%~of~Fe_2O_3~in~the~sample

3 0
3 years ago
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