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VARVARA [1.3K]
3 years ago
15

Is butter solution or colloid or suspension

Chemistry
1 answer:
netineya [11]3 years ago
6 0
Hi , butter is a class of colloids called emulsions , so your answer is colloid.
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Formaldehyde, ch2o, is used as an embalming agent. draw the structure of ch2o including lone pairs.formaldehyde, ch2o, is used a
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<span>Aldehydes and ketones have the C=O grop in their structure Formaldehyde : ......O ......ll ......C ..../....\ ..H....H</span>
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3 years ago
Pb(NO3)2(aq) + KCl(aq) ---- → KNO3(aq) + PbC12(s)
emmasim [6.3K]
If you’re asking to balance the equation then:

Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)

Just remember: the equations at the end is Cl not C12

Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
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2 years ago
What determines crystal size in minerals formed by lava or magma?
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Answer:

When magma cools, crystals form because the solution is super-saturated with respect to some minerals. If the magma cools quickly, the crystals do not have much time to form, so they are very small. If the magma cools slowly, then the crystals have enough time to grow and become large.

Explanation:

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2 years ago
Consider the four free body diagrams. Free-body diagrams are diagrams used to show the relative magnitude and direction of all f
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If you look closely at each of the four diagrams you would be able to conclude that

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3 years ago
Read 2 more answers
The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati
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Answer:  3.72 M

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 = 3.66\times 10^{-3}s^{-1}

t = age of sample = 15.0 minutes

a = let initial amount of the reactant  = 10.0 M

a - x = amount left after decay process = ?

15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}

\log\frac{100}{(a-x)}=1.43

\frac{100}{(a-x)}=26.9

(a-x)=3.72M

The concentration of H_2O_2 in a solution after 15.0 minutes have passed is 3.72 M

7 0
3 years ago
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