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adell [148]
3 years ago
12

How to calculate moles

Chemistry
2 answers:
blondinia [14]3 years ago
8 0

Explanation:

First you must calculate the number of moles in this solution, by rearranging the equation. No. Moles (mol) = Molarity (M) x Volume (L) = 0.5 x 2. = 1 mol.

For NaCl, the molar mass is 58.44 g/mol. Now we can use the rearranged equation. Mass (g) = No. Moles (mol) x Molar Mass (g/mol) = 1 x 58.44. = 58.44 g.

ValentinkaMS [17]3 years ago
6 0

Answer:

Number of moles : Mass (g) ÷ relative atomic mass

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Gemiola [76]

Answer: what is your issue

Explanation:

7 0
2 years ago
A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati
prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

k=\frac{2.303}{48.0}\log\frac{53500}{42520}

k=9.98\times 10^{-2}hr^{-1}

Now we have to calculate the half-life.

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

9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}

t_{1/2}=6.94hr

Therefore, the half life of 28-Mg in hours is, 6.94

8 0
3 years ago
Week 4 learning task 1 science 4th quarter​
cestrela7 [59]

Answer:

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Explanation:

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8 0
2 years ago
Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,
gogolik [260]
<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

6 0
3 years ago
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
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