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Natasha_Volkova [10]
2 years ago
5

How many atoms are in 16.1 G Sr

Chemistry
1 answer:
marysya [2.9K]2 years ago
4 0

Answer: There are 1.11\times 10^{23} atoms of Sr

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{16.1g}{87.6g/mol}=0.184moles

1 mole of Sr contains =  6.023\times 10^{23} atoms of Sr

Thus 0.184 moles of Sr contains =  \frac{6.023\times 10^{23}}{1}\times 0.184=1.11\times 10^{23} atoms of Sr

Thus there are 1.11\times 10^{23} atoms of Sr

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klemol [59]

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

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Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

8 0
3 years ago
The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of
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Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})

= 1401.5 kJ

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We know that, 1kcal = 4.184kJ

So, 1401.5kJ(\frac{1kcal}{4.184kJ})

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.


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