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nikklg [1K]
3 years ago
5

What is the mass of one mole of calcium carbonate

Chemistry
2 answers:
vlada-n [284]3 years ago
6 0
100.08 grams
Use train track method!
valkas [14]3 years ago
3 0
The mass of one mole of calcium carbonate should be 100.087g. Hope this helps!
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Help bc i need it lol
nexus9112 [7]

Answer:

<u>2</u> XE + <u>1</u> AB. --> <u>1</u> AE2 + <u>2</u> XB

Above equation is now balanced

6 0
3 years ago
4) If a person is riding a bike and want to slow down, they use the brakes. This is an example of which of the following?
Vlad1618 [11]

Answer: C A B

Explanation:

4 0
3 years ago
How many moles of AlCl3 are formed by 1.53 moles of CuCl2?
jolli1 [7]

Answer:

1.02 moles of AlCl3

Explanation:

Consider the balanced reaction equation shown below;

2Al(s) + 3CuCl2(aq) -------> 2AlCl3 (aq) + 3Cu(s)

This is the balanced reaction equation for the reaction going on above. Recall that the first step in solving any problem is to accurately put down the chemical reaction equation. This balanced reaction equation always serves a reliable guide in solving the problem at hand.

Given that;

3 moles of CuCl2 yields 2 moles of AlCl3

1.53 moles of CuCl2 yields 1.53 × 2 / 3 =1.02 moles of AlCl3

5 0
3 years ago
As current flows through a system, what happens to the voltage available at each load?
navik [9.2K]

Answer: voltage drops in each resistor ΔU= RI

Explanation: if lamps or other resistor which cause load are in series in

Electric circuit, current I passing circuit is same. Voltage decreases

In every resistor

3 0
3 years ago
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina
liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0
3 years ago
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