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3 years ago

(03.03 MC)

Chemistry

1 answer:

Anit [1.1K]3 years ago

7 0

Answer: C

Explanation: Picture attached.

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What is a colligative property?

V125BC [204]

Colligative properties are properties of solutions that depend on the number of molecules [or ions] in a given volume of solvent and not on the properties (e.g. size or mass) of the compound. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure.

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3 years ago

. What is the mass of 23.56 moles of Glucose ples of Glucose - C6H1206?

GaryK [48]

Answer:

4244.48 g to the nearest hundredth.

Explanation:

The molar mass of Glucose = 6*12.011 + 12*1.008+ 6*15.999

= 180.156.

So 23.56 moles = 180.156 * 23.56 = 4244.48 g

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4 years ago

A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold

Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

$\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}$

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3 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$

So,

Volume of water = mass/density

$\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$$

=4.5 L (Answer)

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4 years ago

What are the chemical formulas for..

andrezito [222]

S4O5
S3O
SeF6
N4S5
CCl9

All numbers should be subscripts

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3 years ago