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2 years ago

Complete the following double replacement reactions b. ZnCO3 + BaCl2

Chemistry

1 answer:

Anna11 [10]2 years ago

7 0

J the same idea that but I’m in gonna right now I’m going out of my work class tomorrow and I will text

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A sample of methane (CH4) has a volume of 25 mL at a pressure of 0.80 atm. What is the volume of the gas at each of the followin

Xelga [282]

Answer:

a. 50ml b.10ml c. 6.097ml d. 190.1 ml

Explanation:

According to Boyle's law

Volume is inversely proportional to pressure at constant temerature

Mathematically

P1V1=P2V2

P1=Initial pressure=0.8atm

V1=Initial volume=25ml

making V2 the subject

at 0.4atm P2=0.4 atm,

V2=25×0.8/0.4

=50ml

at 2 atm V2=25×0.8/2

=10 ml

1mmHg=0.00131579

2500mmHg=3.28 atm

At 3.28 atm,V2=25×0.8/3.28

=6.097 ml

at 80.0 torr

1 torr=0.00131579

80 torr=0.1052 atm

at 0.1048 atm V2=25×0.8/0.1048

=190.1 ml

4 0

2 years ago

In a hot shower or steam room, you look up at the ceiling. What do you observe on the ceiling and why?

Evgen [1.6K]

Steam rises to the top and stays there

3 0

3 years ago

The bond type in CaO is

sergey [27]

CaO is an ionic bond

4 0

3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago

I need help with number 8 please help ASAP.

Rzqust [24]

Answer:

33.33% = 33%

Explanation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

1 mole of MCO3 will produce → 1 mole of CO2

We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %

8 0

3 years ago