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4 years ago

What is the number placed in front of a compound to balance a chemical reaction?

1 answer:

5 0

coefficient: they balance the chemical equation you have to make sure the number is as small as it can. It is also used to convert different compounds to compounds or quantities to quantities.

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How many moles are in 153’pp grams of KClO3

katen-ka-za [31]

Answer:

n = 1.24 moles

Explanation:

Given that,

Mass = 153 grams

Molar mass of KClO₃ = 122.55 g/mol

We need to find the number of moles.

We know that,

No. of moles = given mass/molar mass

So,

$n=\dfrac{153}{122.55 }\\\\n=1.24$

So, there are 1.24 moles in 153 g of KClO₃.

4 0

3 years ago

What device do scientists use to measure radiation

Rufina [12.5K]

Answer:

Geiger counter

Explanation:

Came from Google.

6 0

4 years ago

Read 2 more answers

An electron in an atom is known to be in a state with magnetic quantum number ml=−1

Ulleksa [173]

Explanation:

It is known that the maximum value of ml is equal to the vale of l. But the minimum value of n is as follows.

n = l + 1

where, n = principle quantum number

l = azimuthal quantum number

Values of n can be 1, 2, 3, 4 and so on. Whereas the values of l can be 0, 1, 2, 3, and so on.

Also, "m" is known as magnetic quantum number whose values can be equal to -l and +l.

So, when n = 1 then l = 0 and m = 0.

When n = 2 then l = 1 and values of m will be equal to -1, 0, +1. As it is given that the magnetic quantum number ml = -1. Hence, it is only possible when n = 2.

Thus, we can conclude that the smallest possible value of the principal quantum number n of the state is 2.

5 0

3 years ago

What does Avogadro’s number represent? the number of atoms in 12 grams of the isotope carbon 12 the number of atoms in 120 grams

Julli [10]

The number of atoms in 12 grams of the isotope carbon 12.

4 0

4 years ago

Read 2 more answers

Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g.

AnnZ [28]

Answer:

$m=107.8g$

Explanation:

Hello,

In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h of sleeping:

$E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ$

Then, we compute the energy used that day:

$E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ$

Finally, the mass by considering the consumed fat:

$m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g$

Regards.

4 0

3 years ago