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Sholpan [36]
2 years ago
6

The density of Silver is 10.5 g/mL. How much mass would be present given a silver cube

Chemistry
1 answer:
seraphim [82]2 years ago
3 0

Answer:

m=dxv

= 10.5 x 965 = 10132.5 rounded off to 10100 g

Explanation:

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Speed

Explanation:

The distance travelled by a body per unit time

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Signals that are the same or similar to their original energy source.
dmitriy555 [2]
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6 0
2 years ago
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I need answers to question 1,2,3
sashaice [31]

Answer:

1. 0.125 mole

2. 42.5 g

3. 0.61 mole

Explanation:

1. Determination of the number of mole of NaOH.

Mass of NaOH = 5 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mole of NaOH =?

Mole = mass /molar mass

Mole of NaOH = 5/40

Mole NaOH = 0.125 mole

2. Determination of the mass of NH₃.

Mole of NH₃ = 2.5 moles

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mass of NH₃ =?

Mass = mole × molar mass

Mass of NH₃ = 2.5 × 17

Mass of NH₃ = 42.5 g

3. Determination of the number of mole of Ca(NO₃)₂.

Mass of Ca(NO₃)₂ = 100 g

Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]

= 40 + 2[14 + 48]

= 40 + 2[62]

= 40 + 124

= 164 g/mol

Mole of Ca(NO₃)₂ =?

Mole = mass /molar mass

Mole of Ca(NO₃)₂ = 100 / 164

Mole of Ca(NO₃)₂ = 0.61 mole

6 0
2 years ago
Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
3 0
2 years ago
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An equilibrium is not changed by a change in pressure.<br><br> True<br> OR<br> False
geniusboy [140]
It IS changed. When there is pressure, itll move along to the side of where there are fewer moles of gas in the reaction. Decrease in pressure causes it to shift to the side with more moles of gas. So, false, it IS in fact changed
5 0
2 years ago
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