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3 years ago

A positive nitrite test suggests the presence of a minimum of

1 answer:

7 0

A positive nitrite test suggests the presence of a minimum of 105 microorganisms / mL in your urine. The nitrite test in urine is a method used to detect the presence of microorganisms in the urine. This is a diagnostic test for urinary tract infections. Microorganisms that are the cause of UTI or urinary tract infection produce an enzyme that would change nitrates in urine into nitrite. This nitrite that is formed is the ion that is monitored and measure in a nitrite test. If nitrites are present in the urine, then most probably UTI is present. A positive in this test is called nitrituria.

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What describes a property of matter ability to change into different substances

Brilliant_brown [7]

Answer:

Chemical reaction

Explanation:

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4 years ago

What trend in ionization do you see going up and down?

xxTIMURxx [149]

Depending if you were going up, you would be seeing an increase in relative ionization energies in a particular group on the periodic table, and vice versa, if you go down the group, an decrease in the relative ionization energies.

5 0

3 years ago

The energy produced in the stars results from

faust18 [17]

Answer:

Explanation:

Hydrogen fusion (nuclear fusion of four protons to form a helium-4 nucleus) is the dominant process that generates energy in the cores of main-sequence stars. It is also called "hydrogen burning", which should not be confused with the chemical combustion of hydrogen in an oxidizing atmosphere.

3 0

3 years ago

A binary compound created by reaction of nitrogen and an unknown element E contains 30.46% N and 69.54% E by mass. If the formul

aleksklad [387]

Answer:

15.95

Explanation:

This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.

Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.

In 100 grams of the binary compound we have

30.46 g N

69.54 g E

The number of moles is the mass divided by atomic weight:

mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N

mol E = 65.54 g / A.W E

Thus,

4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18

2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g

So the A.W is 15.94 g/mol which is close the atomic weight of O.

5 0

3 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0

3 years ago