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asambeis [7]
3 years ago
9

What are the two halves of the brain

Chemistry
1 answer:
ch4aika [34]3 years ago
7 0

Answer: left and right hemisphere ya

Explanation:

You might be interested in
If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
8 0
3 years ago
A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2
mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: 2NO \Leftrightarrow N_{2}O_{2}    (fast)

Rate expression for step 1 is as follows.

               Rate = k [NO]^{2}

Step 2: N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O    (fast)

Here, N_{2}O_{2} is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0
3 years ago
How many grams do 6.534e+24 molecules of phosphoric acid weigh?
fredd [130]

Answer:

1,063 grams H₃PO₄

Explanation:

To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).

Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)

Molar Mas (H₃PO₄): 97.99 g/mol

6.534 x 10²⁴ molecules H₃PO₄                       1 mole                         97.99 g
---------------------------------------------  x  -------------------------------------  x  --------------
                                                            6.022 x 10²³ molecules          1 mole

= 1,063 grams H₃PO₄

3 0
2 years ago
A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?
Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is d=\frac{m}{v} . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

d=\frac{10g}{6cm^{3} }

d=1.67\frac{g}{cm^{3} }

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is \frac{g}{cm^{3} } (grams per centimetres cubed).

8 0
3 years ago
Read 2 more answers
Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Clas
JulijaS [17]

Answer:

HCN, weak acid

H⁺, Br⁻, strong acid

Explanation:

Hydrocyanic acid is a weak acid, according to the following equation.

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Thus, it should be written in the undissociated form (HCN).

Hydrobromic acid is a strong acid, according to the following equation.

HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)

Thus, it should be written in the ionic form (H⁺, Br⁻).

7 0
3 years ago
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