3 years ago

Solve the problems in the image above

Chemistry

2 answers:

natulia [17]3 years ago

7 0

I'm sorry idk I'm very sorry

deff fn [24]3 years ago

7 0

Answer:

this is really confusing

Explanation:

Im so sorry but this picture looks so idk im sorry

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Suppose that a fictitious element, x, has two isotopes: 59x (59.015 amu) and 62x (62.011 amu). the lighter isotope has an abunda

professor190 [17]

The atomic mass of the element would simply be equal to the sum of the weighted average of each isotope, that is:

atomic mass = 59.015 amu * 0.717 + 62.011 amu * (1 – 0.717)

<span>atomic mass = 59.863 amu</span>

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3 years ago

Quickly Please!!!!!!

FrozenT [24]

Answer:

$n=4.1mol$

Explanation:

Hello,

In this case, we can use the ideal gas equation:

$PV=nRT$

So we know the temperature, pressure and volume, therefore we can easily compute the required moles as shown below:

$n=\frac{PV}{RT}=\frac{2.0atm*101L}{0.082\frac{atm*L}{mol*K}*600K} \\\\n=4.1mol$

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3 years ago

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Strike441 [17]

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The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

Nimfa-mama [501]

Answer:

$\boxed{\text{254 g}}$

Explanation:

$\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}$

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