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Lelechka [254]
2 years ago
12

HELP DUE TOMMOROW! FAILING CLASS!!! When it would be appropriate to use Radiocarbon dating, Rubidium-Strontium dating, Uranium-L

ead dating, and Potassium-Argon dating. Include if the types of dating are used for organic or inorganic things and the time range they are useful for.
Chemistry
1 answer:
gladu [14]2 years ago
4 0

Answer:

Uhm if you're failing the class why are you still using brainly to get answers... Where's your momma boy?

Explanation:

It izzz what it izzz

You might be interested in
How many moles are in 6.777 x 1023 atoms of CO2. Watch your significant figures. Question 2 options: 1 atoms CO2 1.125 moles CO2
Archy [21]

Answer:

1.13moles

Explanation:

Given parameters:

Number of atoms  = 6.777 x 10²³ atoms

Unknown:

Number of moles  = ?

Solution:

A mole of a substance contains the avogadro's number of particles

       6.02  x  10²³ particles   = 1 mole

      6.777 x 10²³ atoms will contain \frac{ 6.777 x 10^{23}  }{6.02 x 10^{23} }   = 1.13moles

6 0
2 years ago
Formula for Ni2+ and NO3
lyudmila [28]

The formula of a compound is Ni(NO_3)_2

Given:

The ions Ni^{2+} and NO_3^-

To find:

The formula for ions Ni^{2+} and NO_3^-

Solution:

The ion with a positive charge (cation) = nickel ion = Ni^{2+}

The ion with a negative charge (anion) = nitrate ion = NO_3^-

Valency on nickel ion = 2

Valency on nitrate ion = 1

Using the criss-cross method as shown in image

The formula of a compound : Ni(NO_3)_2

The name of a compound is nickel(II) nitrate

Ni^{2+}+2NO_3^-\rightarrow Ni(NO_3)_2

The formula of a compound is Ni(NO_3)_2

Learn more about ions here:

brainly.com/question/24389121?referrer=searchResults

6 0
3 years ago
How many formula units make up 25.8g of magnesium chloride (MgCl2)? Express the number of formula units numerically.
Verdich [7]

Answer:

Explanation:

Kandndkjwnww

8 0
3 years ago
After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
zzz [600]

Answer:

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.

Explanation:

Hi

The percentage of water in the sample is lower than expected.

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:

If part of the sample is splashed from the crucible the mass of water detected will be less.

B. The dehydrated sample absorbed moisture after heating:

If the sample absorbs water after heating the percentage of water would be higher than expected.

C. The amount of the hydrate sample used was too small:

Depending on the sample size, different procedures can be chosen for analysis.

D. The crucible was not heated to constant mass before use:

In many occasions the crucible is heated next to the sample and not in previous form.

E. Excess heating caused the dehydration sample to decompose:

If the sample decomposes during heating, the analysis should be discarded.

success with your homework

4 0
3 years ago
1. How many joules of heat are required to raise the temperature of 750 g of water from 11.0 oC to 19.0 oC?
Leya [2.2K]

Answer:

  1. 25080 J
  2. 146.9 g
  3. 92.58 °C
  4. 0.808 J/g°C
  5. 117.09 g
  6. a. 1708.8 kJ  b.1246.56 kJ
  7. 368.55 kJ
  8. 6.81 kJ
  9. 5.50 grams of methane produces more heat than 5.5 grams of propane.

Explanation:

  1. The specific heat capacity of water=4.18 J/gK

The enthalpy change is calculated using the formula: ΔH=MC∅ where ΔH is the change in enthalpy, M the mass of the substance, C the specific heat capacity of the substance and ∅ the temperature change.

Thus, ΔH= 750g × 4.18 J/gK × (19-11)K

=25080 J

2. Enthalpy change= mass of substance × specific heat capacity of the substance× Change in temperature.

ΔH= MC∅

M= ΔH/(C∅)

Substituting for the values in the question.

M=8750 J/(0.9025/g°C×66.0 °C)

=146.9 grams

3. Enthalpy change =mass × specific heat capacity × Temperature

ΔH= MC∅

∅ = ΔH/(MC)

=6500 J/(250 g × 4.18 J/g°C)

=6.22° C

Final temperature =98.8 °C - 6.22°C

=92.58 °C

4. Specific heat capacity =mass × specific heat capacity × Temperature change.

ΔH=MC∅

C= ΔH/(M∅)

Substituting with the values in the question.

C = 4786 J/(89.0 g×(89.5° C-23°C))

=0.808 J/g°C

5. Heat lost lost copper is equal to the heat gained by water.

ΔH(copper)= ΔH(water)

MC∅(copper)=MC∅(water)

M×0.385 J/g°C× (75.6°C- (19.1 °C+5.5°C))=100.0g×4.18 J/g°C×5.5 °C

M=(100.0g×4.18J/g°C×5.5°C)/(0.385 J/g°C×51 °C)

=117.09 grams.

6 (a). From the equation 1 mole of methane gives out 890.4 kJ

There fore 2 moles give:

(2×890.4)/1= 1780.8 kJ  

(b) 22.4 g of methane.

Number of moles= mass/ RFM

RFM=12 + 4×1

=16

No. of moles =22.4 g/16g/mol

=1.4 moles

Therefore 1.4 moles produce:

1.4 moles × 890.4 kJ/mol=

=1246.56 kJ

7. From the equation, 2 moles of aluminium react with ammonium nitrate to produce 2030 kJ

Number of moles = mass/RAM

Therefore 9.75 grams = (9.75/26.982) moles of aluminium.

=0.3613 moles.

If 2 moles produce 2030 kJ, then 0.3613 moles produce:

(0.3631 moles×2030 kJ)/2

=368.55 kJ

8. From the equation, 4 moles of ammonia react with excess oxygen to produce 905.4 kJ of energy.

Number of moles= mass/molar mass

RMM= 14+3×1= 17

Therefore 0.5113 grams of ammonia = (0.5113 g/17g/mole) moles

= 0.0301 moles

If 4 moles produce 905.4 kJ, then 0.0301 moles produce:

(0.0301 moles×905.4 kJ)/4 moles

=6.81 kJ

9. From the equations, one mole of methane produces 890 kJ of energy while one mole of propane produces 2043 kJ.

Lets change 5.5 grams into moles of either alkane.

Number of moles= Mass/RMM

For propane, number of moles= 5.5g/ 44.097g/mol

=0.125 moles

For methane number of moles =5.5 g/ 16g/mol

=0.344 moles

0.125 moles of propane produce:

0.125 moles×2043 kJ/mol

=255.375kJ

0.344 moles of methane produce:

0.344 moles× 890 kJ/mol

= 306.16kJ

Therefore, 5.5 grams of methane produces more heat than 5.5 grams of propane.

6 0
3 years ago
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