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4 years ago

Equilateral triangle What’s the answer

Chemistry

2 answers:

lianna [129]4 years ago

6 0

Answer:

Explanation:

3 sides are the same

sesenic [268]4 years ago

3 0

No answer choice??!????????????

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

Is cellulose living material?

Natasha_Volkova [10]

Yes cellulose is alive because it contains cells

3 0

3 years ago

What are different type of plastids explain them?

BARSIC [14]

Answer:

Explanation:

There are 3 types of plastids :-

1) Chloroplasts:- The green plastids which contain chlorophyll pigments for photosynthesis.

2) Chromoplasts:-The coloured plastids for pigment synthesis and storage.

3) Leucoplasts:- The colourless plastids for monoterpene synthesis found in non- photosynthetic parts of the plants.

They are of three types:-

a) Amyloplasts- stores starch.

b) Proteinoplasts- stores proteins.

c) Elaioplasts- stores fats and oils.

6 0

3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

8 0

3 years ago

At 393 K, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 478 K, assuming constant volume?

True [87]

Answer:

About 1.301 atm

Explanation:

The formula that you should is PV=nRT, where P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the universal gas constant, and T stands for temperature in Kelvin. Since the volume, number of moles, and universal gas constant don't change, you don't need to worry about them.

1.07V=393nR

PV=498nR

P=1.301 atm. Hope this helps!

7 0

3 years ago