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11 months ago

a technician mixes 80 ml of a 5% solution with 10 ml of water. what is the final percentage strength of the solution prepared?

1 answer:

5 0

A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

given that :

8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL

therefore, 80 × 5 = 10 × x %

x % = 40 %

Therefore, the final percentage strength of the solution is 40 %

Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

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The diagram below shows part of the rock cycle. (6 points)

Anna35 [415]

Answer:

Igneous Rock

Explanation:

Assuming this is a cycle, the volcanic eruption would lead back to rock B; rocks formed by volcanic eruptions are considered Igneous.

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2 years ago

national defense, currency, post office, foreign affairs, & interstate commerce. b. charter local governments, education, pu

katen-ka-za [31]

Answer And Explanation:

Option C is correct.

Lend and borrow money, taxation, law enforcement, charter banks and transportation.

Some of the powers that were mentioned in the other options that weren't concurrent powers (that is, they belong to either the state government alone or the federal government alone) & disqualified them from being the answer include:

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3 years ago

The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 + X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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3 years ago

How many grams of acetylene are produced by adding water to 5.00 g CaC2?

Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

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2 years ago

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