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wlad13 [49]
3 years ago
9

True or False: Rubbing a balloon on your head for a longer amount of time will increase the attractive force of the balloon.​

Chemistry
1 answer:
Andreyy893 years ago
4 0

Answer:

no

Explanation:

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A gas occupies a volume of 50.0 mL at 100K and 630 mmHg. At what temperature would
rodikova [14]
PV / T = P'V' / T'

V = V'

P / T = P' / T'

P = 630 mmHg
T = 100 K
P' = 1760 mmHg
T' = ?

630 / 100 = 1760 / T'

T' = 1760 / 6,3

T' = 279,36 K

T' ≈ 280 K
3 0
3 years ago
Write a paragraph explaining the effect of surface area on the rate of reaction.
nexus9112 [7]
(For a bit of context I will use the reaction between HCl and Mg as an example)

The larger the surface area of the magnesium metal, the more particles are exposed to collide with the aqueous HCl particles to cause the reaction to occur. This increases the frequency per second of collisions, speeding up the rate of reaction.

The effect of a catalyst is to reduce the minimum collision energy which allows the reaction to happen. This does not increase the number of collisions per second, but increases the percentage of successful collisions, which consequently causes the rate of reaction to increase .

I have drawn diagrams showing the effect of surface area, but there isn't really a meaningful diagram that I know of to show the impact of a catalyst (at least not at GCSE level).


6 0
3 years ago
How many grams of fosforic acid (H3PO4) are required to prepare 500 mL of a 0.2 M solution?
navik [9.2K]

Answer:

Mass = 9.8 g

Explanation:

Given data:

Molarity of solution = 0.2 M

Volume of solution = 500 mL

Number of grams of phosphoric acid = ?

Solution:

First of all we will convert the volume milliliter to litter.

500 mL × 1 L/1000 mL

0.5 L

Molarity = moles of solute / volume in litter

0.2 M =  number of moles / volume in litter

Number of moles = 0.2 mol/L × 0.5 L

Number of moles = 0.1 mol

Number of grams:

Number of moles = mass/molar mass

Mass = number of moles × molar mass

Mass = 0.1 mol × 98 g/mol

Mass = 9.8 g

7 0
3 years ago
What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) => 2
zavuch27 [327]

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, \frac{6}{635}\times 44.7=0.423mol of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

7 0
4 years ago
Photosynthesis
3241004551 [841]
CONVERTS GLUCOSE INTO ENERGY AND WATER
7 0
4 years ago
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