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Vikki [24]
3 years ago
5

The amount of kinetic energy an object has depends on

Chemistry
2 answers:
Vladimir79 [104]3 years ago
7 0

it depends on the mass & speed of the moving object.^-^

Alex777 [14]3 years ago
3 0
How much it has to drop and how heavy it is. Hope this is what you're looking for:)
You might be interested in
Decreasing order of C-C bond length is 1) ethene 2)ethyne 3) benzene 4) ethane
Likurg_2 [28]

Answer:

4>3>1>2.

That is, the C-C bond length in ethane > benzene > ethene > ethyne.

Explanation:

The C-C bond in ethane is single, the C-C bond in ethene is double and the C-C bond in ethyne is triple. As the number of bonds between a C-C increases, the length of the bond decreases with an increase in strength. This explains why the C-C bond length in ethane > ethene > ethyne. For benzene, all the C-C bonds in the aromatic compound has been found to have an identical length of 1.40 Å, compared to ethane (1.54Å), ethene (1.34Å) and ethyne (1.20Å). Hence the trend in bond lengths: ethane > benzene > ethene > ethyne.

6 0
3 years ago
The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the te
cricket20 [7]

Answer:

A. P₂ / P₁ = 2

B. P₂ / P₁ = 1.1

Explanation:

A. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 46 K

Final temperature (T₂) = 92 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/46 = P₂/92

Cross multiply

46 × P₂ = P₁ × 92

Divide both side by P₁

46 × P₂ / P₁ = 92

Divide both side by 46

P₂ / P₁ = 92 / 46

P₂ / P₁ = 2

B. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K

Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/308.4 = P₂/342

Cross multiply

308.4 × P₂ = P₁ × 342

Divide both side by P₁

308.4 × P₂ / P₁ = 342

Divide both side by 308.4

P₂ / P₁ = 342 / 308.4

P₂ / P₁ = 1.1

5 0
3 years ago
Which of the following is the atomic number of an alkali metal?​
xxMikexx [17]

Answer:

what are the options??

Explanation:

6 0
3 years ago
What parts are present in every chemical equation?
WITCHER [35]
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
5 0
2 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
3 years ago
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