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Veseljchak [2.6K]
3 years ago
13

If you guys don’t mind can you explain to me why u think it’s either A, b, c, d plz and thank you

Chemistry
1 answer:
sveticcg [70]3 years ago
8 0

Answer: answer is D

Explanation:

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What is the atomic radius <br> of mercury?
cricket20 [7]

The answer is 155 pm, hope this helps!

4 0
2 years ago
Someone please answer this...
kupik [55]

Answer:

5446.8 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 50 g

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Specific heat capacity (C) = 0.89 J/gºC

Heat (Q) required =?

Next, we shall determine the change in the temperature. This can be obtained as follow:

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 192.4 – 70

ΔT = 122.4 °C

Finally, we shall determine the heat required to heat up the block of aluminum as follow:

Mass (M) = 50 g

Specific heat capacity (C) = 0.89 J/gºC

Change in temperature (ΔT) = 122.4 °C

Heat (Q) required =?

Q = MCΔT

Q = 50 × 0.89 × 122.4

Q = 5446.8 J

Thus, the heat required to heat up the block of aluminum is 5446.8 J

5 0
2 years ago
If earth did not rotate what would happen to the hlobal wind
Readme [11.4K]
It would probably stop moving. Earth has motion and we do do. without the world moving, there would be No wind at all.
8 0
2 years ago
What is the % composition of Carbon in Chromium (iii) Carbonate
photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

The percent composition of Carbon is thus 12.7 %.

8 0
1 year ago
A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1%
Otrada [13]
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)

6 0
3 years ago
Read 2 more answers
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