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Ostrovityanka [42]
3 years ago
15

How many grams of water are frozen if 6,346 Joules of energy was given

Chemistry
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

19g

Explanation:

Given parameters;

Amount of energy  = 6346J

Unknown:

Mass of water frozen = ?

Solution:

This is a phase change and a latent heat was involve;

        H  = mL

H is the quantity of heat

m is the mass

L is the latent heat of fusion of water  = 335J/g

So;

        6346  = mass x  335

    Mass = 19g

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Mendeleev put elements in groups while creating the periodic table. What important role do groups play in the structure of the p
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Answer:

Groups show similar chemical properties about the elements

Explanation:

The periodic table is divided into groups and periods. Groups are vertical while periods are horizontal.

All elements in the same group possess the same number of outermost shell electrons. This number of outermost shell electrons controls the reactivity of the elements in the group.

Hence, groups show similar chemical properties of the elements in them, Groups separate elements into families which assist in the speedy study of elements.

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Which is an "enumerated" power of Congress?
fredd [130]

Answer:

In this section, we consider how several of the enumerated powers of Congress under the original Constitution have been interpreted. The Congress shall have Power To lay and collect Taxes, Duties, Imposts and Excises, to pay the Debts and provide for the common Defence and general Welfare of the United States.

6 0
3 years ago
Read 2 more answers
Chemistry Lab Question: Filtration
tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

6 0
3 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
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Nady [450]
Answer/explanation:

Product identifier – the brand name, chemical name, common name, generic name or trade name of the hazardous product.
Initial supplier identifier – the name, address and telephone number of either the Canadian manufacturer or the Canadian importer.
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