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ss7ja [257]
3 years ago
5

If you were converting from milli- to centi- units, would you move the decimal point to the left or the right

Chemistry
1 answer:
yan [13]3 years ago
6 0

Answer:

Left

Explanation:

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A salt contains only magnesium and one of the halide ions. A 0.0776-g sample of the salt was dissolved in water, and an excess o
sdas [7]

Answer:

The formula of the magnesium halide is MgF₂

Explanation:

All halides, X, produce a salt with Mg with the formula:

MgX₂

<em>-There are 2 moles of the halide ion per mole of Mg-</em>

<em />

With the mass of the MgSO₄ we can find moles of magnesium sulfate  = Moles Mg.

With moles of Mg we can know the moles of the halide -1 mole Mg = 2 moles of Halide-

And we can find the mass of Mg in the 0.0776g sample. Subtracting we can find the mass of the halide and, with the mass and moles of the halide we can find its molecular weight and its identity:

<em>Moles MgSO₄ -Molar mass: 120.366g/mol- = Moles Mg:</em>

0.150g * (1mol / 120.366g) = 1.2462x10⁻³ moles Mg

<em>Moles halide:</em>

1.2462x10⁻³ moles Mg * 2 = <em>2.4924x10⁻³ moles Halide</em>

<em>Mass Mg -Molar mass: 24.305g/mol:</em>

1.2462x10⁻³ moles Mg * (24.305g / mol) = 0.0303g Mg

<em>Mass halide:</em>

0.0776g - 0.0303g Mg = <em>0.0473g</em>

<em>Molecular weight of the halide:</em>

0.0473g / 2.4924x10⁻³ moles =

18.98g/mol

This molecular weight is the molecular weight of Fluoride ion, F⁻,

<h3>The formula of the magnesium halide is MgF₂</h3>
6 0
3 years ago
A buffer is prepared by mixing 50.3 mL of 0.183 M NaOH with 128.8 mL of 0.231 M acetic acid. What is the pH of this buffer
kvv77 [185]

Answer:

A buffer system can be made by mixing a soluble compound that contains the conjugate ... 10.0 grams of sodium acetate in 200.0 mL of 1.00 M acetic acid.

Explanation:

7 0
2 years ago
Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantites.
Bingel [31]

Answer:

A. Kilo , K, multiplication by 1000

B. Centi, c

C. Deci, d

D. Mili, m

E. Mega, M

F. Micro, u

7 0
3 years ago
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0
1 year ago
How many hydrogen bonds are between cytosine and guanine
Dahasolnce [82]

Answer:

3

Explanation:

5 0
2 years ago
Read 2 more answers
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