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3 years ago

What is calcium's matching element

Chemistry

1 answer:

vodomira [7]3 years ago

8 0

Answer: Explanation:

Calcium (Ca), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table. It is the most abundant metallic element in the human body and the fifth most abundant element in Earth's crust.

...

Calcium.

atomic number 20

oxidation state +2

electron configuration 1s22s22p63s23p64s2

Explanation:

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If you use 19 mL of 0.100 M KMnO4, how many moles have you used? The answer has to be in four decimal places (do not forget the

Scorpion4ik [409]

1)

Volume = 19 / 1000 = 0.019 L

n = M * V

n = 0.100 * 0.019

n = 0.0019 moles
__________________________

2)

M₁ * V₁ = M₂* V₂

5.0 * V₁ = 1.0 * 152

5.0 V₁ = 152

V₁ = 152 / 5.0

V₁ = 30.4 mL
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<span>hope this helps!</span>

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The density of gold is 19.3 g/cm3. what volume would 131.2 g of gold have?

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3 years ago

What is usually composed of two or more minerals

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3 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

8 0

3 years ago

What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0

3 years ago