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sergiy2304 [10]
1 year ago
13

Approximately 50% of our bone is chemically calcium phosphate, Ca3(PO4)2

Chemistry
1 answer:
yuradex [85]1 year ago
4 0
Yea ion know i just want sun point lol o m g is it rekutten ;)
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If an object is not moving its kinetic energy is zero.<br><br> True<br> False
Svetach [21]

Answer: This is true because kinetic energy depends on speed. If there's no speed, then there is no kinetic energy.

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What is simplest level at which life may exist
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Determine the specific heat of copper from the fact that 38 g of 80°C copper are needed to raise the temperature of 15 g of wate
mars1129 [50]
From the equation q=mCΔT, set the q of copper = to q of water,

So --- mCΔT(copper)=mCΔT(water).

mass (Cu - copper) = 38g
mass (H2O - water) = 15g
C (H2O) = 4.184 J/g*C
ΔΤ (H2O) = 33-22 = 11*C
ΔΤ (Cu) = 33-80 = -47*C (the final temp is the same for both materials - thermal equilibrium)
C (Cu) = ?

So --- 38(-47)C[Cu]=15(4.184)(11)
     --- C[Cu]=690.36/(-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)
5 0
3 years ago
CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST
jekas [21]

Explanation:

2. 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

First, we need to find the number of moles of CO_2 at 300K and 1.5 atm using the ideal gas law:

n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}

=2.0\:\text{mol}\:CO_2

Now use the molar ratios to find the number of moles of ethane to produce this much CO_2.

2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)

=1.0\:\text{mol}\:C_2H_6

Finally, convert this amount to grams using its molar mass:

1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)

=30.1\:g\:C_2H_6

3. 3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2

Convert 75 g Zn into moles:

75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn

Then use the molar ratios to find the amount of H2 produced.

1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2

Now use the ideal gas law PV=nRT to find the volume of H2 produced at 23°C and 4 atm:

V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}

=8.9\:\text L\:H_2

8 0
2 years ago
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