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3 years ago

10 points pls help!!

Chemistry

2 answers:

adell [148]3 years ago

5 0

Answer:

Question 16: I'm not sure but most likely #3

Question 17: #4

Explanation: Hope this Helps

Trava [24]3 years ago

5 0

Answer:

A for first one and second one is A

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At a pressure of 505 kPa, the volume of a gas is 12.00 mL. Assuming the temperature remains constant,

Sergio [31]

Answer:

757.5 kPa

Explanation:

Boyle's law states that p1 * v1 = p2 * v2

v1 = 12 mL

v2 = 8 mL

p1 = 505 kPA

p2 = ?

(505)(12) = (p2)(8)

6060 = (p2)(8)

p2 = 757.5

8 0

3 years ago

Describe the structure and bonding in graphite.

yan [13]

Answer:

In graphite, each carbon atom is covalently bonded to 3 other carbon atoms. ... These extra electrons are delocalised, or free to move, in the area between layers of carbon atoms. As these electrons are free to move they are able to carry charge and thus graphite can conduct electricity.

Explanation:

3 0

3 years ago

Read 2 more answers

In the balanced chemical reaction for the combustion of acetylene (used in welding torches), determine at standard temperature a

Tpy6a [65]

Answer:

8L of CO2

Explanation:

The equation for the reaction is given below:

2C2H2 + 5O2 —> 4CO2 + 2H2O

From the equation above,

5L of O2 produced 4L of CO2.

Therefore, 10L of O2 will produce = (10 x 4)/5 = 8L of CO2

Therefore, 8L of CO2 is produce

8 0

4 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

Which of the following is a product of photosynthesis? Carbon dioxide Glucose Water Xylem

serious [3.7K]

The answer is glucose.
Hope it helps!

5 0

3 years ago