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Luden [163]
3 years ago
5

A compound has a pH of 3 what is the [H+] concentration?

Chemistry
1 answer:
Allisa [31]3 years ago
8 0

Answer: 0.001 M

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution. Mathematically, pH is expressed as

pH = -log(H+)

where H+ is hydrogen ion concentration

So, 3 = -log(H+)

(H+) = Antilog (-3)

(H+) = 0.001

Thus, [H+] concentration is 0.001 M

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PLEASE HELP ME !!!!!!!
love history [14]

Answer:

it would be 5,045

Explanation:

because it is closer to 5,000. pls correct me if wrong

5 0
2 years ago
Check my answer?
rusak2 [61]
Yes that is correct because hypotheses should be testable meaning it can be proven wrong and it has to be reasonable meaning it should make sense. 
8 0
3 years ago
Read 2 more answers
Write the quantities 3 x 10^-4 cm and 3 x 10^4 km in ordinary notation
Goshia [24]
To write quantities in ordinary notation, you need to notice the power of the exponent.
If the power is positive, then you move the decimal point to the right by the number the power in exponent tells you.
If the power is negative, then you move the decimal point to the left by the number the power in exponent tells you.

We have:
3 x 10^-4 : the power of exponent is negative, therefore, we will move the decimal point 4 places to the left.
3 x 10^-4 = 0.0003 km

3 x 10^4 : he power of exponent is positive, therefore, we will move the decimal point 4 places to the right.
3 x 10^4 = 30000 km
6 0
3 years ago
Please answer this Q:
Anarel [89]
D. container four because of the water
3 0
2 years ago
How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?
nika2105 [10]

Answer:

m_{C_8H_{18}}=85.67gC_8H_{18}

Explanation:

Hello there!

In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}}   \\\\m_{C_8H_{18}}=85.67gC_8H_{18}

Regards!

6 0
3 years ago
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