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Mice21 [21]
4 years ago
7

How many formula units in 3.56 g magnesium phosphate

Chemistry
1 answer:
Rudik [331]4 years ago
6 0

Answer:

1.5 of mag 1 phos 1 oxygen

Explanation:

hope this helps well sorry i don't really know lol i tried

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3 years ago
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If you purchased 0.610 μCi of sulfur-35, how many disintegrations per second does the sample undergo when it is brand new?Expres
ASHA 777 [7]

Answer:

the sample undergo 1.67 \times 10^5  disintegrations per second.

Explanation:

Given:

The amount of sulphur purchased is 0.671 μCi

To find:

disintegrations per second = ?

Solution:

Some of the conversions are

1 Ci = 3.8 \times 10^{10} Bq

1 rad = 0.01 Gy

1Gy = 1 j/kg tissue

1 rem = 0.01 Sv

1Sv = 1 j/Kg

Using these conversions,

The decay rate of this sample is calculated as

0.671 \mu C i \times \frac{1 \mu C i}{10^{6}} \mu C i \times \frac{0.671 \times 10^{10}}{1 \mu C i}

= 1.67 \times 10^5 disintegrations per second

7 0
4 years ago
Someone please help will mark as brainliest
Bas_tet [7]

Answer:

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Explanation:

7 0
3 years ago
a) Diamond and graphite are two different forms of pure elemental carbon with densities of 3.51 g/cc and 2.25 g/cc respectively.
dybincka [34]
Density can be calculated using the following rule:
density = mass / volume
Therefore,
volume = mass / density

For diamond:
we have mass = 0.5 grams and density = 3.51 g / cm^3
Substituting in the rule, we can calculate the volume of diamond as follows:
volume = 0.5 / 3.51 = 0.14245 cm^3

For graphite:
we have mass = 0.5 grams and density = 2.25 g / cm^3
Substituting in the rule, we can calculate the volume of graphite as follows:
volume = 0.5 / 2.25 = 0.2223 cm^3
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3 years ago
What is the molarity of a potassium triiodide solution, ki3(aq), if 30.00 ml of the solution is required to completely react wit
oee [108]

The molar concentration of the KI_3 solution is 0.0833 mol/L.

<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>

Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)

<em>Step 2</em>. Calculate the <em>moles of I_3^(-) </em>

Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)

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5 0
4 years ago
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