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3 years ago

What are the numerical values defined by STP?

1 answer:

8 0

Standard temperature and pressure are standard sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data. ... Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

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Non motile in an sentences

allsm [11]

Answer:

Motile bacteria have flagella, while nonmotile bacteria do not.

Explanation:

4 0

3 years ago

What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7

mariarad [96]

The empirical formula of hydrocarbon is CH2

The molecular formula of the hydrocarbon is C6H12

<u><em>Explanation</em></u>

Hydrocarbon is made up of carbon and hydrogen

<h3><u><em> </em></u>Empirical formula calculation</h3>

Step 1: find the moles CO2 and H2O

moles =mass/molar mass

moles of CO2 = 6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol = 0.15 moles

Step 2: Find the moles ratio of Co2:H2O by diving each mole by smallest mole(0.15)

that is for CO2 = 0.15/0.15 =1

For H2O = 0.15/0.15 =1

therefore the mole ratio of Co2 : H2O = 1:1 which implies that 1 mole of Co2 and 1 mole of H2O is formed during combustion reaction.

From the the law of mass conservation the number of atoms in reactant side must be equal to number of atoms in product side

therefore since there 1 atom of C in product side there must be 1 atom of C in reactant side.

In addition there is 2 H atom in product side which should be the same in reactant side.

From information above the empirical formula is therefore = CH2

Molecular formula calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n = 84 g/mol

n = 6

multiply the each subscript in CH2 by 6

Therefore the molecular formula = C6H12

5 0

3 years ago

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

6 0

3 years ago

What is the weighted average of a nail in the sample data given?

Natalka [10]

The weighted average of the nail in accordance with the given data is 11.176g.

<h3>How to calculate weighted average?</h3>

Weighted average is an arithmetic mean of values biased according to agreed weightings.

The weighted average of the nail in the image above can be calculated by multiplying the decimal abundance with the mass of the nail, then summed up as follows;

Weighted average = (decimal abundance × mass 1) + (decimal abundance × mass 2)

Weighted average = (0.12 × 3.3) + (0.88 × 12.25)

Weighted average = 0.396 + 10.78

Weighted average = 11.176g

Therefore, 11.176g is the weighted average of the nail

Learn more about weighted average at: brainly.com/question/28042295

#SPJ1

5 0

1 year ago

How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?

DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

$2*n_{acid}=n_{base}$

And in terms of concentrations and volumes:

$2*M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the volume of acid:

$V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL$

Best regards.

5 0

3 years ago