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3 years ago

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What is the molar of mass of copper (II) sulfate CuSOu?

1 answer:

Aleksandr-060686 [28]3 years ago

8 0

191.546,,,,,,,,,,,,,,,,,,,

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A intramuscular medication is given at 5.00mg/kg of body weight. What is the dose in grams for a 180-lb patient?

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0.408 gram for 180-lb patient

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3 years ago

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Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older?

ch4aika [34]

Answer:

Fossil B

Explanation:

Fossil B is older because it's later is at the bottom

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3 years ago

1929 cm3 of gas is how many L of gas

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1.929 Liters of gas

Explanation:

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3 years ago

The mass defect for the formation of lithium-6 is 0.0343 g/mol. The binding energy for lithium-6 nuclei is ________ kJ/mol. Ente

Sholpan [36]

<u>Answer:</u> The binding energy for lithium-6 nuclei is 3.09 E+11

<u>Explanation:</u>

Binding energy is defined as the energy which holds the nucleus together. It is basically the product of mass defect and the square of the speed of light.

This energy is calculated by using Einstein's equation, which is:

$E=\Delta mc^2$

where,

E = Binding energy of the atom

$Delta m$ = Mass defect = 0.0343g/mol = $0.0343\times 10^{-3}kg/mol$ (Conversion factor: $1kg=10^3g$ )

c = speed of light = $3\times 10^8m/s$

Putting values in above equation, we get:

$E=0.0343\times 10^{-3}kg/mol\times (3\times 10^8m/s)^2$

$E=3.09\times 10^{14}J/mol=3.09\times 10^{11}kJ/mol$ (Conversion factor: $1kJ=10^3J$ )

Hence, the binding energy for lithium-6 nuclei is 3.09 E+11

7 0

3 years ago

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solutio

Zigmanuir [339]

<u>Answer:</u> The concentration of original sulfuric acid solution is 1.62 M

<u>Explanation:</u>

Let the original concentration of sulfuric acid be 'x' M

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=xM\\V_1=25.00mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$x\times 25.00=M_2\times 250.0\\\\M_2=\frac{x\times 25.0}{250}=\frac{x}{10}$

Now, to calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=\frac{x}{10}M\\V_1=10.00mL\\n_2=1\\M_2=0.1790M\\V_2=18.07mL$

Putting values in above equation, we get:

$2\times \frac{x}{10}\times 10.00=1\times 0.1790\times 18.07\\\\x=\frac{1\times 0.1790\times 18.07\times 10}{2\times 10.00}=1.62M$

Hence, the concentration of original sulfuric acid solution is 1.62 M

5 0

3 years ago