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ahrayia [7]
3 years ago
9

Can y’all please help me

Chemistry
1 answer:
Scrat [10]3 years ago
8 0
The measure is the same. :)
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Student Exploration: Limiting Reactants Vocabulary: chemical equation, chemical formula, chemical reaction, coefficient, limitin
SIZIF [17.4K]

Answer:

30 hot dogs

Explanation:

It is given that :

There are 4 packets of eight wieners, i.e. 4 x 8 = 32 wieners

There are 3 bags of ten buns, i.e. 3 x 10 = 30 buns

One hot dogs need 1 bun and 1 wiener to make a hot dog.

There are 30 buns, so 30 hot dogs can be made out by using all the 30 buns and the 30 wieners out of the 32 wieners.

Therefore, 30 hot dogs.

And the number of extra wieners left = 32 - 30 = 2 wieners.

6 0
3 years ago
Where, approximately, is the negative pole on each of these molecules?
Leni [432]

Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.

Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.

5 0
3 years ago
Read 2 more answers
Need help with the 1st one
lutik1710 [3]

If I’m right this should be the answer

7 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

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

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
3 years ago
When balancing a chemical equation can you adjust the number that is subscripted to a substance formula?
Ugo [173]
No, you can't do that
3 0
3 years ago
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