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photoshop1234 [79]
3 years ago
7

When I googled how to balance this chemical equation K2CrO4+NaHSO3+H2O=NaHSO4+KOH+Cr(OH)3 they gave me the balanced form of 2K2C

rO4+3NaHSO3+5H2O=3NaHSO4+4KOH+2Cr(OH)3. But I don't know how to get this because to me it would seem unbalanced.
Chemistry
1 answer:
galina1969 [7]3 years ago
3 0
Whatever is on one side of the equation needs to be the same as the other. So K2CrO4+NaHSO3+H2O=NaHSO4+KOH+Cr(OH)3 you need to balance the equation between the = to make a balanced equation.
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Among the alkali earth metals, the tendency to react with other substances
tatyana61 [14]
Answer: option <span>A) increases from bottom to top within the group.

Explanation:

</span>It is a known trend that the metallic character of the elements increase from let to right and from top to bottom.

The greater the metallic character the greater the reactivity of the metal.

So, the elements of the columns 1 and 2 are the most reactive metals and among them the elements at the bottom are yet more reactive.

<span>The higher reactivity of the metals that are lower in the periodic table is attributed to the greater total number of electrons.

The greater the total number of electrons the more reactive the metals as their outermost electrons (the valence electrons which are those that react) are located further from the nucleus and therefore they are held less strongly, which makes them react more easily.</span>
5 0
2 years ago
Read 2 more answers
4
matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

5 0
2 years ago
1 message Which process results in the increase in entropy of the universe? the cooling of a hot cup of coffee in room temperatu
andreyandreev [35.5K]

Answer:

Melting of snow

Evaporation of water from desk

Explanation:

Processes that increase the entropy of the universe are those processes that have an increased disorderliness. We should note that there are three principal states of matter which are the liquid, gas and solid. The gaseous state is the most disorderly while the solid is the least disorderly.

Now. We can see that the cooling of a hot cup of coffee is a process that needs or leads to a loss in temperature which obviously decreases disorderliness of the universe.

The melting of snow however is a process that leads to an increase in the disorderliness of the universe. It entails moving from the solid state to the liquid state. It tends to move to a more disordered state indicating an increase in the entropy of the universe.

The evaporation of water from the desk is quite similar to that above. Hence since we are moving from the liquid to the gaseous state via evaporation, we can state that the entropy of the universe has increased since we have moved from a state with a lesser degree of disorderliness to a state that is more disordered I.e from liquid to gaseous state.

3 0
2 years ago
When sedimentary rock sits for years and years will it build up more layers and. Make fossils
KengaRu [80]
Sedimentary rocks don't necessarily make fossils. Although when animals die on top of the rocks and are never moved, blown away, etc they can get trapped under the growing rock. Which is why fossils are often found in sedimentary rocks.
6 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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