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poizon [28]
3 years ago
5

Please help me..what is carbon 1/12​

Chemistry
1 answer:
tatuchka [14]3 years ago
6 0

Answer:

It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units. ... In this scale 1 atomic mass unit (amu) corresponds to 1.660539040 × 10−24 gram.

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!Chemistry help!<br> How many moles are in a 1500 mL solution with a 1.2 M?
Ray Of Light [21]

Answer:

69 ;)

Explanation:

7 0
3 years ago
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What is the daughter nucleus of pt-199 after a beta decay?
Setler79 [48]

B. At-200

In beta decay the atomic number increases by 1

6 0
3 years ago
Balanced symbol equation for reaction between sodium carbonate solution and dilute sulphuric acid?
Dmitry [639]

Answer:

The molecular equation for the reaction betweensodium carbonate and sulfuric acid is: 1. Na2CO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l) N a 2 C O 3 ( a q ) + H 2 S O 4 ( a q ) → N a 2 S O 4 ( a q ) + C O 2 ( g ) + H 2 O ( l ) .

Explanation:

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3 years ago
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A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

8 0
3 years ago
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

7 0
3 years ago
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