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Oduvanchick [21]
4 years ago
10

How much heat is required to raise the temperature of 88.5 g of water from its melting point to its boiling point?

Chemistry
1 answer:
k0ka [10]4 years ago
5 0
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The chemical formula for strontium sulfite is SrSO3.<br><br><br> TRUE<br><br><br> FALSE
VLD [36.1K]
Valency is the number of electrons lost or gained by an atom to attain an stable configuration. Valency is important when writing the formula of chemical compounds in chemistry. Strontium has a valency of  2 while sulfite ion (radicle) has a valency of 2. Therefore, the chemical formula of strontium sulfite is written as SrSO3.
3 0
3 years ago
HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5
marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction  

HA (aq) ---> H⁺ (aq) + A⁻ (aq)  

Ka's value  

\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔  (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}

[H₃O⁺]=2.86 x 10⁻⁶ M

5 0
3 years ago
If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of
maw [93]

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

4 0
4 years ago
WILL MARK BRAINLIEST IF YOU ANSWER WITHIN 5 MINUTES
tensa zangetsu [6.8K]

Answer:

This is and ADDITION REACTION

Explanation:

Because your putting a compound and an element together

6 0
3 years ago
Read 2 more answers
What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer
Zepler [3.9K]
d_{1}=\frac{m}{V_{1}}\\\\&#10;V_{1}=395cm^{3}\\&#10;d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\&#10;d_{2}=\frac{m}{V_{2}}\\\\&#10;m=150,1g\\&#10;d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}
8 0
4 years ago
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