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Sedaia [141]
3 years ago
9

3. The density of a substance is 1.80 g/ml. What is the mass of 0.32 L of the substance in grams?

Chemistry
2 answers:
s2008m [1.1K]3 years ago
5 0

Answer:

576 g

Explanation:

as mass = density × volume

Anika [276]3 years ago
4 0
<h3><u>Answer</u> :</h3>

★ Density of substance is defined as the mass per unit volume.

  • d = M/V

where,

d denotes density

M denotes mass

V denotes volume

Given that :

  • Density = 1.80 g/mL
  • Mass = 0.32 L

By substituting the values,. we get

⭆ M = d × V

⭆ M = 1.80 × 0.32

⭆<u> M = 0.576g</u>

<h3>Hope It Helps!</h3>
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The effective nuclear charge is an innate property of a specific element. It is the pull of force that an electron feels from the nucleus. It is related to the valence electron by the equation: Z* = Z-S, where Z* is the effective nuclear charge, Z is the atomic number and S is the shielding constant.

For the following elements in the choices, these are their values of Z*:

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5 0
3 years ago
A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.
lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

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The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
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Answer:

Approximately 2.46\; \rm mol.

Explanation:

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\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

3 0
3 years ago
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