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oee [108]
3 years ago
12

For an object to sink

Chemistry
1 answer:
AfilCa [17]3 years ago
5 0

Answer:

For and object to sink, it must have more density than the liquid in which it is placed. For example, if you have a glass of water and a metal spoon, the spoon will sink because it is both heavier than the water, therefore having more density.

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Balance the following chemical equation : C7H6O3 + C4H6O3 —> C9H8O4 + H2O
Nana76 [90]

Answer:

7C7H6O3 + 8C4H6O3 = 9C9H8O4·H2O

Explanation:

6 0
3 years ago
What is the concentration of H+ ions at a pH = 11?
natta225 [31]

Answer:

\huge 1 × {10}^{-11} \: \: M

Explanation:

The pH of a solution can be found by using the formula

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

Since we are finding the H+ ions we find the antilog of the pH

So we have

11 =  -  log({H}^{+})  \\ {H}^{+} =  {10}^{ - 11}

We have the final answer as

1 × {10}^{-11} \: \: M

Hope this helps you

6 0
3 years ago
The reaction of hydrogen gas with oxygen gas is what type of reaction precipitation single replacement
Oksi-84 [34.3K]
The reaction of hydrogen gas (H2) with oxygen gas (O2) is a COMBINATION or SYNTHESIS reaction, because multiple substances combine to form fewer substances. Here, the two gases form one substance, water (H2O):
2H2 + O2 -- > 2H2O
5 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Sodium hydroxide, NaOH; sodium phosphate, Na3PO4; and sodium nitrate, NaNO3, are all common chemicals used in cleanser formulati
Sholpan [36]
Start by writing out molar mass of each component: 

1) NaOH= 22.99+16+1.01 --> 40 

2) Na₃PO₄= (22.99)3 + 94.97 + 64 --> 227.94

3) NaNO₃= 22.99 + 14 + 48 --> 84.99 

Now compare the mass of sodium with the rest of the substances: 

1) (22.99/40)100%= 57.475%

2) (68.97/227.94)/100%= 30.25% 

3) (22.99/84.99)/100%= 27.05%

As seen, the ordered list is: Sodium hydroxide, sodium phosphate and sodium nitrate. 

Hope I helped :) 
8 0
4 years ago
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