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expeople1 [14]
2 years ago
6

4.8 L sample of helium gas contains 0.22 moles of helium - How many additional moles of herum gas must be added to the sample to

obtain a volume of 6-4L?​
Chemistry
1 answer:
yawa3891 [41]2 years ago
7 0

Answer:

   

Explanation:

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The time necessary for the decay of one- half sample of a radioactive substance is a _____. mass defect daughter half life nucli
Mazyrski [523]

Answer: half life

Explanation: Radioactive decay follows first order kinetics and the time required for the decay of a radioactive material is calculated as follows:

t=\frac{2.303}{k}\hspace{1mm}log\frac{x}{a}

t= time required

k= disintegration constant

x= amount of substance left after time t

a= initial amount of substance

when one half of the sample is decayed, one half of the sample remains and t can be represented as t_{1/2}

at t= t_{1/2}, x=\frac{a}{2}

t_{1/2}=\frac{2.303}{k}\hspace{1mm}log\frac{a/2}{a}

t_{1/2}=\frac{0.693}{k}

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3 years ago
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3 years ago
An acid, like HCI, would...
frosja888 [35]

Answer: melt

Explanation:

6 0
3 years ago
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The boiling point of methanol is 64 7°C. Its melting point is -976°C. At room temperature (-25°C), methanol is in which state?
marissa [1.9K]

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.

The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.

At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults

3 0
2 years ago
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A 20.00-mL sample of a weak base is titrated with 0.0568 M HCl. At the endpoint, it is found that 17.88 mL of titrant was used.
nata0808 [166]

Answer: 0.0508mL

Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.

therefore concentration of the base is 1.0156/20 = 0.0508 mL

7 0
3 years ago
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