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diamong [38]
2 years ago
9

an effervescent tablet is added to a test tube of water, producing a solution that fizzes and releases gas bubbles. The total ma

ss of the substances and the test tube were recorded before and after the tablet was added to the test tube of water.
Chemistry
1 answer:
Alona [7]2 years ago
7 0

Answer:

The mass remains same after and before the reaction

Explanation:

This is because of law of conservation of mass

Which states that

  • In a chemical reaction mass is neither created nor destroyed,it remains conserved .

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Which pure substance is made of only one kind of atom?
aleksley [76]

Answer:

the answer is D i believe. i am not shure but im like 70% sure

7 0
3 years ago
What is difference in reaction of O2 with metal and nonmetal?​
aksik [14]

Answer:

Metal more reactive than non metal

4 0
2 years ago
The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
Rus_ich [418]

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

7 0
3 years ago
2. How much heat is released when 432g of water cools down from 710C to 180C?
nadya68 [22]
The heat released by the water when it cools down by a temperature difference AT

is Q = mC,AT

where

m=432 g is the mass of the water

C, = 4.18J/gºC

is the specific heat capacity of water

AT = 71°C -18°C = 530

is the decrease of temperature of the water

Plugging the numbers into the equation, we find

Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J

and this is the amount of heat released by the water.
7 0
3 years ago
How are scientific questions answered?
dem82 [27]

Answer:

B .Through testing a theory about the physical world

Explanation:

7 0
3 years ago
Read 2 more answers
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