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mina [271]
3 years ago
9

KOH+HBr-KBr+H2O which is the base in this reaction​

Chemistry
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

KOH is the base in this reaction

Explanation:

HBr is the acid in this reaction.

Hope this help you. Have a nice day !

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A chemist wants to produce 12.00 grams of barium sulfate by reacting a .6000 M BaCl2 solution with excess H2SO as show in the re
worty [1.4K]
Find the moles of BaSO4 first. Then since we know it's a one to one ratio from barium chloride to barium sulfate we can just solve for liters.
<span>First you need to find the moles BaSO4 , and the you will require to find barium sulfate in liters.

</span>12.00gBaSO4 / 233.31 grams per mole

=.05141moles

Molarity=moles/liters

Hence,

Liters=.05141moles/.6Molarity
=.85 liters
7 0
3 years ago
Read 2 more answers
How many millimeters of 5 M NaCl are required to prepare 1500 mL of 0.002 M NaCl?
timurjin [86]

Answer:

The correct answer is 0.6 mL.

Explanation:

We use the mathematical expression:

Ci x Vi = Cf x Vf

Where Ci is the initial concentration (5 M); Cf and Vf refers to final concentration (0.002 M) and final volume (1500 mL). With the given data, we calculate the initial volume (Vi):

Vi = (Cf x Vf)/Ci = (0.002 M x 1500 mL)/(5 M) = 0.6 mL

Therefore, we need 0.6 mL of 5 M NaCl to prepare the solution with the requested dilution.

3 0
3 years ago
If 3.5 g of element X reacts with 10.5 g of element Y to form the compound XY.
irina [24]

Answer:

The percentage by mass of element X is 25 %.

The percentage by mass of element Y is 75 %

Explanation:

X + Y ⇒ XY

3.5 g of element X + 10.5 g of element Y = 14g in total

⇒Element X : 3.5g / 14g = 0.25 ⇒ x 100 % = 25 %

⇒Element Y : 10.5 / 14 = 0.75 ⇒ x 100 % = 75 %

The percentage by mass of element X is 25 %.

The percentage by mass of element Y is 75 %

8 0
3 years ago
How many moles of gold are equivalent to 1.204 × 1024 atoms?
LUCKY_DIMON [66]

Explanation:

It is known that the value of Avogadro's number is 6.022 \times 10^{23}.

Therefore, calculate moles of gold atoms as follows.

    Moles of gold atom = \frac{given number of atoms}{Avogadro's number}

                                    = \frac{1.204 \times 10^{24}}{6.022 \times 10^{23}mol^{-1}}

                                    = 0.199 \times 10 mol

                                    = 1.99 mol

Thus, we can conclude that 1.99 mol of gold atoms are equivalent to 1.204 \times 10^{24} atoms.

7 0
3 years ago
Read 2 more answers
Carbon dioxide non examples? Please and catalyst non examples?
PilotLPTM [1.2K]

Answer:

I guess you mean by non element example. Non element examples have more than one elements. Carbon dioxide is a non element example since caebon dioxide has 2 oxygen atoms + 1 carbon atom = carbon dioxide.

I think catalyst is also a non element example because catalyst is inorganic which means that it is not a living thing. Since catalyst is in brass and brass is a non element example, I think catalyst is also a non element example.

Hope that helps, thank you !!

5 0
3 years ago
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