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Paladinen [302]
3 years ago
15

Strontium phosphate, Sr3(PO4)2, is a crystalline substance used in medicine and industry. How many phosphorus atoms are represen

ted in the formula for Sr3(PO4)2? a) 2b) 3c) 4d) 8
Chemistry
1 answer:
dexar [7]3 years ago
7 0

Answer: option A. 2

Explanation: in the formula Sr3(PO4)2, the 2 behind (PO4) is affecting both P and O4. It means that we have P2 in the formula

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Calculate the number of moles in a 2.98 g sample of aluminum.
Ksju [112]
Molar mass <span>Aluminium ( Al) = 26.98 g/mol

1 mole ------------ 26.98 g
? mole ----------- 2.98 g

moles Al = 2.98 x 1 / 26.98

moles Al = 2.98 / 26.98

= 0.110 moles

hope this helps!

</span>
3 0
2 years ago
Species developed harder shelled eggs to raise them on land, this is to adapt to the changes in the enviorment during the Cenozo
mario62 [17]
False would be the answer
6 0
3 years ago
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Which subscripts would properly complete the formula unit Al_N_?
MakcuM [25]
I believe that it would be Al1N1.
4 0
3 years ago
Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe
stealth61 [152]
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So 
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
5 0
2 years ago
if 5430 J of energy is used to heat 1.25 L of room temp. water (23.0 °C) whats the final temp of the water?
skelet666 [1.2K]

<span>We can use the heat equation,
Q = mcΔT </span>

 

<span>Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature difference (°C).</span>


Density = mass / volume


The density of water = 0.997 g/mL

<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>

<span>                                                                   = 1246.25 g</span>


Specific heat capacity of water = 4.186 J<span>/ g °C.</span>


Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

      5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
               T = 1.04 °C + 23 °C
               T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.
4 0
3 years ago
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