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3 years ago

ERGENT IM TIMED!!!!!!!!!!!

Chemistry

2 answers:

torisob [31]3 years ago

3 0

Answer:

HBr is the strongest and its bond is what which makes it the strongest acid.

Explanation:

This acid is dangerous and highly poisonous than other acids. It can disunite instantly. And later then it forms the suspension.

Li_{2}OLi

O and LiBr is not merely an acid and henceH_{2}SH

S and HBr are the acids, amongst which HBr forms the strong acid component.

The strongest nature is due to the formation of hydronium ions.

The Hydrogen sulphide acid does not release hydronium ions on suspension.

Flura [38]3 years ago

3 0

Answer:

HBr is the strongest and its bond is what which makes it the strongest acid.

Explanation:

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Write any five example of radical with their velencies

garri49 [273]

Answer:

Terms in this set (13)

Hydroxide. OH.

Nitrate. NO3.

Ammonium. NH4.

Bicarbonate/Hydrogen Carbonate. HCO3.

Bisulphate. HSO4.

Chlorate. ClO3.

<h2>hope it's correct</h2>

7 0

3 years ago

What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100

Kisachek [45]

What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100

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4 years ago

How does the density of a Hershey bar compare to the density of the a half bar?How does the density of a Hershey bar compare to

Lostsunrise [7]

IDensity is calculated by mass divided by volume. As you half the Hershey bar, mass and volume change by the same proportion. Therefore, density is the same.

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3 years ago

A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat

nexus9112 [7]

<u>Answer:</u> The given sample of water is not safe for drinking.

<u>Explanation:</u>

We are given:

Concentration of fluorine in water recommended = 4.00 ppm

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of fluorine in water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Mass of fluorine = $0.152mg=0.152\times 10^{-3}g$ (Conversion factor: 1 g = 1000 mg)

Mass of water = 5.00 g

Putting values in above equation, we get:

$\text{ppm of fluorine in water}=\frac{0.152\times 10^{-3}}{5}\times 10^6\\\\\text{ppm of fluorine in water}=30.4$

As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.

Hence, the given sample of water is not safe for drinking.

6 0

3 years ago

Natural gas burns in air to form carbon dioxide and water, releasing heat. CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔHrxn = -802.3 kJ.

Olenka [21]

Answer:

1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g) ,\Delta H_{rxn} =-802.3 kJ$

1) Minimum mass of methane required to raise the temperature of water by 21.0°C.

Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 45.0 g\times 4.18 J/g^oC\times 21.0^oC$

Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

$=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol$

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to raise the temperature of water by 26.0°C.

Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC$

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

$=\frac{5.434 kJ}{802.3 kJ}=0.006773 mol$

Mass of 0.006773 moles of methane :

0.006773 mol × 16 g/mol= 0.108 g

Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

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4 years ago