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2 years ago

An excess of hydrogen ions in the body fluids can have disastrous results because

2 answers:

4 0

Excess hydrogen in the body through carbonic acid is known to affect blood. In the blood we have hemoglobin which is a red blood cell protein that allows red blood cells to carry oxygen but when hydrogen is introduced in the blood, of causes some of the hemoglobin to release oxygen, causing oxygen deprivation in major organs. Even 1 minute of oxygen deprivation can cause severe damage to the brain, liver, kidneys, heart, stomach, and every other major organ.

natita [175]2 years ago

3 0

excess hydrogen ions can break chemical bonds

, can change the shape of large complex molecules, rendering them nonfunctional and can disrupt tissue function.

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Choose all the answers that apply. Lakes are recharged by _____. 1.evaporation 2.precipitation 3.rivers 4.transpiration 5.ground

sweet-ann [11.9K]

2. Precipitation, because when it rains the water that was taken when it was evaporated is replaced and it fills the lake back up

6 0

2 years ago

Read 2 more answers

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a

geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0

2 years ago

What is the mass in grams of 7.5 x 10^15 atoms of nickel?

satela [25.4K]

Atomic mass Ni = 58.69 a.m.u

58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms

58.69x (7.5x10¹⁵) / 6.02x10²³

=> 7.31x10⁻⁷ g

6 0

2 years ago

Read 2 more answers

How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?

lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

N₂ + 3 H₂ ------—> 2 NH₃

1 mol 3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

28 grams of N₂ react with 6 g of H₂

if 28 grams of N₂ produces 6 g of H₂ so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

28 g of N₂ ≅ 6 g of H₂

X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

X g of N₂ = 28 g x 14.2 g / 6 g

X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

N₂ + 3 H₂ ------—> 2 NH₃

6 g 2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

6 g of H₂ ≅ 2 mol of NH₃

14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

X mol of NH₃= 14.2 g x 2 mol / 6 g

X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of NH₃

Molar mass of NH₃ = 14 + 3(1)

Molar mass of NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

mass in grams = 4.73 mole x 17 g/mol

mass in grams = 80.41 g

mass of NH₃= 80.41 g

3 0

2 years ago

Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam dist

Snezhnost [94]

Answer:

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,

Partial pressure = mole fraction * total pressure

Mole fraction of water is,

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.

8 0

2 years ago