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morpeh [17]
3 years ago
5

What happens to the speed of molecules in water vapor when it condenses to a form of liquid​

Chemistry
1 answer:
weeeeeb [17]3 years ago
3 0

The speed of the molecules decrease.

When water condenses, the temperature of the water decreases. When the temperature of an molecule decreases, the speed of that molecule also decreases.

When water condenses, the speed of the molecules decreases.

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What is the percentage error of length measurement of 0.229cm if the correct value is 0.225cm
Mars2501 [29]

Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.

So first, we take our measured value, .299 cm, minus our known value, .225 cm.

.299 cm - .225 cm=.004 cm

Next, we divide that by our known value

\frac{.004}{.225}=.0177777778

Finally, multiply your answer by 100

.0177777778 x 100= 1.77777778 %

Round to three significant figures, and you're done.

=1.78 % error



5 0
4 years ago
Read 2 more answers
Please help me with this chem question, I’ll mark you brainiest if it’s right. There’s multiple answers for this one.
Kazeer [188]

Answer:

Option A. KCl (aq)

Option D. Mg(OH)₂(s

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

MgCl₂(aq) + KOH(aq) —>

In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

KOH(aq) —> K⁺(aq) + OH¯(aq)

MgCl₂(aq) + KOH(aq) —>

Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)

MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)

Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)

Thus, option A and D gives the correct answer to the question.

8 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
Which body system processes food into a useable source of energy?
anastassius [24]

Answer:

Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.

4 0
3 years ago
Read 2 more answers
How many moles of glucose can be produced
Arturiano [62]

Answer:

moles of glucose

<u>2.3166 moles of glucose</u>

<u></u>

Explanation:

The balance reaction for the formation of glucose is :

6CO_{2}+6H_{2}O\rightarrow C_{6}H_{12}O_{6}+6O_{2}

here , CO2 = carbon dioxide

H2O = water

C6H12O6 = glucose

O2 = Oxygen

According to this equation :

6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2

We are asked to calculate the mole of Glucose from carbon dioxide.

So,

6 mole of CO2  produce = 1 mole of C6H12O6

1 mole of CO2 will produce =

\frac{1}{6} moles of glucose

13.9 moles of CO2 will produce :

\frac{1}{6}\times 13.9

=2.3166 moles of glucose

Note : first , Always calculate for one mole (By dividing)

. After this , multiply the answer with the moles given.

Always write the substance whose amount is asked(glucose) to the right hand side

5 0
4 years ago
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