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2 years ago

Hhhhhhhhhhhheeeeeeeeeeeellllllllllllllllllpppppppppppppppp

Chemistry

2 answers:

shtirl [24]2 years ago

5 0

Answer:

They have good hearing

They have special adaptations that help them to be successful hunters

<h2>hope it helps</h2>

brainliest please

insens350 [35]2 years ago

4 0

Answer:

They have keen eyesight.

They have special adaptations that help them to be successful hunters.

Add-on:

i hope this helped at all. sorry if its incorrect.

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What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the

Maksim231197 [3]

Answer:

81.59%

Explanation:

4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert 107.50 g of NH₃ into moles, using its molar mass:

107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃:

6.32 mol NH₃ * $\frac{4molNO}{4molNH_3}$ = 6.32 mol NO

Then we convert 6.32 moles of NO to grams, using its molar mass:

6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we calculate the percent yield:

154.70 g / 189.60 g * 100% = 81.59%

8 0

3 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

If the pOH of a cesium hydroxide solution is known to be 4.00. what is the (OH)? *Please round your answer to the appropriate nu

Snowcat [4.5K]

Answer:

i going to be aniston i would say take a gess

4 0

2 years ago

What are the answers for this question Don't just answer for points

Kisachek [45]

Answer:

i cant see it very much i dont think no one can

Explanation:

5 0

3 years ago

Read 2 more answers

Discuss Newland octave law and it's rejection

Delvig [45]

It's where chemical elements are organized based on atomic weight. Also, like/similar physical and chemical properties in each interval of 7 elements.
It was a silly reason it was rejected. It was rejected because it's octaves were too similar to those in music.
I hope this helps!

7 0

3 years ago