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2 years ago

* Write ionic equation, including state symbols for the reaction of

Chemistry

1 answer:

svetoff [14.1K]2 years ago

7 0

Answer:

Ag+ (aq) + Cl- (aq) --> AgCl (s)

Explanation:

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The genetic material of a prokaryotic cell is found in the

Taya2010 [7]

Answer:

The genetic material of a prokaryotic cell is found in the cytoplasm.

Explanation:

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3 years ago

Food moves down into the stomach by:

alexandr1967 [171]

Explanation:

Muscle contraction..

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3 years ago

Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.

bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b) 213.8 °C= 416.84°F

c) 180 °C= 453.15 °K

d) 315 °K= 107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K= -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

Fahrenheit a Celsius: $C=\frac{F-32}{1.8}$

Celsius a Fahrenheit: °F= °C*1.8 + 32
Celsius a Kelvin: °K= °C + 273.15
Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
Fahrenheit a Kelvin: $K=\frac{F-32}{1.8} + 273.15$

Entonces se obtiene:

a) 72 °F= $\frac{72-32}{1.8}$ =22.22 °C

b) 213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= $\frac{1750-32}{1.8} + 273.15$ = 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0

3 years ago

Creatinine, , is a by-product of muscle metabolism, and creatinine levels in the body are known to be a fairly reliable indicato

kherson [118]

Answer:

Molality = 0.0862 mole/kg

Explanation:

Molality = (number of moles of solute)/(mass of solvent in kg)

Number of moles of solute = (mass of Creatinine in the blood sample)/(Molar mass of Creatinine)

To obtain the mass of creatinine in 10 mL of blood. We're told that 1 mg of Creatinine is contained in 1 decilitre of blood.

1 decilitre = 100 mL

1 mg of Creatinine is contained in 100 mL of blood

x mg of Creatinine is contained in 10 mL of blood.

x = (1×10/100) = 0.1 mg = 0.0001 g

Molar mass of Creatinine (C₄H₇N₃O) = 113.12 g/mol

Number of moles of Creatinine in the 10 mL blood sample = (0.0001/113.12) = 0.000000884 moles

Mass of 10 mL of blood = density × volume = 1.025 × 10 = 10.25 mg = 0.01025 g = 0.00001025 kg

Molality of normal creatinine level in a 10.0-ml blood sample = (0.000000884/0.00001025)

Molality = 0.0862 moles of Creatinine per kg of blood.

Hope this Helps!!!

7 0

3 years ago

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Yuliya22 [10]

Via the half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where A is the amount initially and final amount after some time has passed, t is the time elapsed, h is the half life time and so:

$A_{final}=45(\frac{1}{2})^\frac{140}{35} = 2.8125g$

Therefore there will be 2.8125 grams left after 140 seconds.

7 0

4 years ago