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3 years ago

What is the total Mass of Reactants in Alpha Decay

Chemistry

2 answers:

sweet [91]3 years ago

8 0

Answer:

Calcite Calcium carbonate

Calgon Calcium hexametaphosphate

Calomel Mercurous chloride

Carbolic acid Phenol

Explanation:

⭐ sorry ⭐

Naddik [55]3 years ago

6 0

Ey do.pe bro if u keep posting free pts ur gonna get ur account deleted or banned.

Btw what did u mean by trolling ppl ?

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Ur my new bestie lm.ao

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Hillsdale and Cordera are the same distance from the equator, and both are near the ocean. Is Cordera warmer, colder, or the sam

kozerog [31]

Answer:

Cordera is colder than Hillsdale.

Explanation:

They may be at a similar latitude. But you also have to consider the currents nearby the location.

The prevailing wind closer to the equator which carries warm current hits Hillsdale which is why it follows the coastline of the continent. However, for Cordera prevailing winds carrying cold currents hits the continent and follows or hits the coastline of Cordera.

8 0

2 years ago

a brick measures 25cm by 12cm by 13cm. What is the volume of the brick in cm cubed? how many milliliters of water would this bri

julia-pushkina [17]

The volume is 3900cm^3 and the bricks density is 3900 ml

5 0

3 years ago

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc

jeka94

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

$T_i$ = 21.5°C

$C_{calorimeter}$ = 6.15 kJ/°C

$T_f$ = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

4 0

3 years ago

The reason an increase in substrate concentration does not increase the reaction rate indefinitely is because

scoundrel [369]

Answer:

Increasing substrate concentration also increases the rate of reaction to a certain point. Once all of the enzymes have bound, any substrate increase will have no effect on the rate of reaction, as the available enzymes will be saturated and working at their maximum rate.

3 0

2 years ago

PLEASE HELP!!! HOW DO I DO THIS??

Effectus [21]

4.7 is the answer!

Mark me brainiest

7 0

1 year ago