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3 years ago

If you start with 5 molecules of O2 in the reaction 2Mg + O2 2 MgO, how much Mg will you need

Chemistry

1 answer:

hichkok12 [17]3 years ago

4 0

2Mg + O₂ ---> 2MgO

2 atoms Mg --- 1 molecules O₂
X atoms Mg --- 5 molecules O₂
X = 2×5
X = 10 atoms Mg

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Question 2 (1 point)

Alexxandr [17]

Answer:

a mole have have wrong becouse i understand

4 0

2 years ago

Calculate each of the following quantities:<br> (a) Total number of ions in 38.1 g of SrF₂

adelina 88 [10]

The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.

Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.

We have to calculate the total number of ions in 38.1 g of .

The molar mass of SrF₂ = 125.62 g/mol

The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles

Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles

Total moles of ions in 0.30329 moles of

= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions

We know that,

1.0 mole of ions = 6.023 x 10²³ ions

Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions

Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions

To learn more about ions, refer to the link:

brainly.com/question/14295820

#SPJ4

8 0

1 year ago

The pH of a basic solution is 9.77. What is [OH⁻]?

uysha [10]

The [OH⁻] of the solution is 5.37×10⁵ M.

This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.

To calculate the hydroxy ion [OH⁻] concentration we use the formula below.

Note:

pH+pOH = 14

pOH = 14-pH
pOH = 14-9.77
pOH = 4.27

Formula:

[OH⁻] = 1/ $10^{pOH}$ ................. Equation 1

Given:

pOH = 4.27

Substitute the value into equation 1

[OH⁻] = 1/ $10^{4.27}$
[OH⁻] = 5.37×10⁵

Hence, The [OH⁻] of the solution is 5.37×10⁵ M.

Learn more about hydroxy ion concentration here: brainly.com/question/17090407

7 0

2 years ago

Read 2 more answers

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

3 0

3 years ago

What is missing from the temperature and volume<br> graph shown at right?

Anna11 [10]

X axis hopes this helps

8 0

2 years ago

Read 2 more answers