Does hydrogen builds all the acids??

The ph of a solution for which [OH]= 1.0 x 10^-4 is? A 10 B 14 C 11 D 9 E 7

zhenek [66]

To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

pH= -log (1.0x10-10)= 10

answer is A

8 0

3 years ago

What happens in the process of beta decay?

Andrej [43]

Answer:

A neutron transforms into a proton and an electron.

Explanation:

i took the test got an 100%

6 0

3 years ago

Why does opening the air valve of a tire at a constant temperature decrease the pressure

Shtirlitz [24]

The number of molecules decrease

8 0

3 years ago

How many atoms of oxygen are present in 7.51 grams of glycine with formula C₂H5O2N?

Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0

1 year ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

Does hydrogen builds all the acids??​

Does hydrogen builds all the acids??