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Genrish500 [490]
3 years ago
7

I’m only having a problem with this number that’s all if anyone could help I would really appreciate it! Thank you!

Chemistry
1 answer:
Masja [62]3 years ago
7 0

Answer:D.

Explanation:i used photo math. It is really helpful when I'm doing math homework etc...

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A Geiger counter is used to measure
Fantom [35]
It detects ionizing radiation<span> such as alpha particles, beta particles and gamma </span>rays<span>using the ionization effect produced in a </span>Geiger<span>–Müller tube; which gives its name to the instrument.</span>
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3 years ago
draw the lewis structure for CO2, H2CO3, HCO3-, and CO3 2-.Rank these in order of increasing attraction to water molecules. Expl
gavmur [86]

Answer:

The structures are attached in file.

Hydrogen bonding and intermolecular forces is the reason for ranks allotted.

Explanation:

In determining Lewis structure, we calculate the overall number of valence electrons available for bonding.  Making carbon (the least electronegative atom) the central atom in the structure, we allocate valence electrons until each atom has achieved stability.

In order of decreasing affinity to water molecules:

CO_{3}^{2-}  > HCO_{3} ^{2-} > H_{2} CO_{3}

This is due to the fact that the CO_{3}^{2-}will accept protons more readily than the bicarbonate ion, HCO_{3} ^{2-}. Carbonic acid, H_{2} CO_{3} will not accept any more protons, hence it is the least attractive to water molecule, even though soluble.

3 0
3 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
3 years ago
Copper oxide is a: (Multiple Choice)
Triss [41]

Answer:

The answer is 5. Compound

7 0
3 years ago
Read 2 more answers
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

MM= 49.2 g/mol

3 0
3 years ago
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