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2 years ago

Is water in a kettle boiling a chemical change

Chemistry

2 answers:

Andru [333]2 years ago

7 0

Yes it is a chemical change

Luden [163]2 years ago

6 0

When a kettle is boiling you are able to see the chemical reaction, from the stem leaving the kettle, so yes
Hope this helped : )

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Which of the following is an example of positive acceleration? Q A boy running from his house around the block and ending up bac

Natalija [7]

The answer is A track star speeding up as he sprints to the finish line.

8 0

3 years ago

Is oxygen gas (O2) a compound or an element? Is hydrogen gas (H2) a compound or an element?

Alinara [238K]

They're both elements

4 0

3 years ago

Sulfur trioxide, so3, is made industrially in enormous quantities by combining oxygen and sulfur dioxide, so2. what amount (mole

DedPeter [7]

The answer is 37.5 moles

The explanation:

1- when we have the mass of SO3 in Kg so, first we have to convert it to grams:

mass = 3 * 1000 = 3000 g

2- we need to get the molar mass of SO3 :

molar mass of SO3 = 32 + (16*3)

= 80 g/mol

3- then we can use this formula to get number of moles:

moles = mass / molar mass

= 3000 g / 80 g/mol

= 37.5 moles

6 0

3 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

3 years ago

Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho

nikdorinn [45]

To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]

HOBr = 0.50 M
KOBr = 0.30 M = OBr-

 HOBr + H2O <-> H+ + OBr- 
I 0.50 - 0 0.30 
C -x x x
---------------------------------------------
E(0.50-x) x (0.30+x) 

Assuming that the value of x is small as compared to 0.30 and 0.50 

Ka = 2.0 x 10^-9 = x (0.30) / 0.50) 

x = 3.33 x 10^-9 = H+
pH = 8.48

3 0

3 years ago