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Jobisdone [24]
3 years ago
10

What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

1.196 M NaOH

Explanation:

Molarity = moles/Volume (L)

moles NaOH = mass NaOH/MM NaOH = 12/40.01 = 0.299 moles NaOH

Volume solution = 250 mL = 0.250L

M = 0.299/0.250=1.196 M NaOH

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`Let's just consider the concept of stoichiometry without any confusing chemicals. Here is a generic equation: 2A + 6B ? 3C. If
lesya [120]

Answer:

Option D. 5.5

Explanation:

The equation is this:

2A + 6B  ⇒  3C

With the amounts that we were given, let's determine which is the <em>limting reactant</em>

2 A reacts with 6 B

4 A will react with ( 4 .6)/2 = 12B

I have 11 B, so the limiting is B

6 B react with 2 A

11 B will react with (11 .2 )/6 =3.66 A

I have 4 A, so A is the excess.

6 B produce 3 C

11 B will produce ( 11 .3)/6 = 5.5C

7 0
3 years ago
A 16.4 g sample of calcium chloride is dissolved in enough water to make 750.0 mL of solution. Calculate the molarity of calcium
EastWind [94]
SIMP FOR LEVI AKRAMEN
7 0
3 years ago
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
3 years ago
Chem<br> Equations<br> Balance and Classify each of the following equations into:<br> Combination reaction, Decomposition Reacti
Arlecino [84]

Answer :

(A) The balanced chemical reaction will be:

2KBr(aq)+Cl_2(g)\rightarrow 2KCl(aq)+Br_2(l)

This reaction is a single replacement reaction.

(B) The balanced chemical reaction will be:

This reaction is a double displacement reaction.

(C) The balanced chemical reaction will be:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

This reaction is a combination reaction.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of an element present on reactant side must be equal to the product side.

Part (A):

The balanced chemical reaction will be:

2KBr(aq)+Cl_2(g)\rightarrow 2KCl(aq)+Br_2(l)

This reaction is a single replacement reaction in which the most reactive element displaces the least reactive element from its solution.

Part (B):

The balanced chemical reaction will be:

CaBr_2(aq)&#10;+H_2SO_4(aq)\rightarrow  CaSO_4(s)+2HBr(g)

This reaction is a double displacement reaction in which a positive cation and a negative anion of two reactants exchange their places to form two new products.

Part (C):

The balanced chemical reaction will be:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

This reaction is a combination reaction in which the two atoms combine to form a larger molecule.

3 0
3 years ago
PLEASEEE HELPP ME
lina2011 [118]

Answer:

natural resource management

Explanation:

the management of natural resources such as land, water, soil, plants and animals with a particular focus on how management affects the quality of life for both present and future generations.

Did this help? Or do you need me to find a better answer

4 0
3 years ago
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