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3 years ago

PLS HURRY MY TEST IS TIMED

Chemistry

2 answers:

Rainbow [258]3 years ago

7 0

Answer:

Track down where the fertilizer was sold and who bought it

Explanation:

I, myself, took this exam.

aliina [53]3 years ago

6 0

Answer:

Extract DNA from the fertilizer to identify its origin.

Explanation:

Barnes find out that fertilizer was used in the explosive material. Barnes should now identify the origin of this type of fertilizer so that its root can be determined. When the origin is unable to be determined then Barnes should keep the traces of the fertilizer and try to identify the manufacturer.

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Is this molecule polar or nonpolar?

lutik1710 [3]

The answer is nonpolar

3 0

3 years ago

True/False Indicate whether the sentence or statement is true or false. The number 1 234 000 in scientific notation is equal to

Shalnov [3]

False bc 1.234 x 10^5 is 123400

8 0

3 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

4 years ago

What is the percent composition HCN?

kupik [55]

Answer:

see calculations in explanation

Explanation:

percent = part/total x 100%

part = ∑ atomic mass of element

hydrogen = 1.008 amu (atomic mass units)
carbon = 12.011 amu
nitrogen = 14.007 amu

total = ∑ molecular mass of compound

= H amu + C amu + Namu

= 1.008 amu + 12.011 amu + 14.007 amu

= 27.026 amu

%H = (1.008amu/27.026amu)100% = 3.730%

%C = (12.011amu/27.026amu)100% = 44.442%

%N = (14.007amu/27.026amu)100% = 51.827%

Check results ∑%values = 100%

3.730% + 44.442% + 51.827% = 99.999% ≅ 100%

7 0

3 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

4 0

4 years ago