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Archy [21]
3 years ago
9

Please please do it because i need an answer.

Chemistry
1 answer:
Natali [406]3 years ago
3 0

Answer:

Positive and Negative electric charge

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How much heat must your body transfer to 500.0g of water to heat it from 25.0°C to body temperature, 37.0°C?
shtirl [24]

Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of water = 500.0 g

c = specific heat of water = 4.18J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 37.0^oC

Now put all the given value in the above formula, we get:

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

Q=25080J=25.1kJ

Therefore, the heat your body transfer must be, 25.1 kJ

3 0
3 years ago
Calculate the mass of beryllium (Be) needed to completely react with 18.9 g nitrogen gas (N2) to produce Bez N2, which is the on
choli [55]

Answer:

C = 18.29 g  

Explanation:

Given data:

Mass of beryllium needed = ?

Mass of nitrogen = 18.9 g

Solution:

Chemical equation:

3Be + N₂    →    Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 18.9 g/ 28 g/mol

Number of moles = 0.675 mol

Now we will compare the moles of nitrogen and Be from balance chemical equation.

                     N₂        :       Be        

                       1          :       3

                  0.675       :      3/1×0.675 = 2.03 mol

Number of moles of Be needed are 2.03 mol.

Mass of Beryllium:

Mass = number of moles × molar mass

Mass = 2.03 mol ×   9.01 g/mol

Mass = 18.29 g  

4 0
3 years ago
An element has an atomic number of 80. How many protons and electrons are in an atom of the element?
Nadya [2.5K]
80 protons,121 Neutrons, and 80 electrons the element is Mercury
6 0
3 years ago
If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?
Y_Kistochka [10]

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

5 0
3 years ago
If the half-life of C-14 is 5730 years, how much C-14 would be left from a 5000 year-old sample that started as 150 grams?
serious [3.7K]
<h3>Answer:</h3>

81.9 grams

<h3>Explanation:</h3>

From the question we are given;

  • Half-life of C-14 is 5730 years
  • Original mass of C-14 (N₀) = 150 grams
  • Time taken, t = 5000 years

We are required to determine the mass left after 5000 years

  • Using the formula;
  • N = No(1/2)^t/T​, where N is the remaining mass, N₀ is the original mass, t is the time taken and T is the half-life.

   t/T = 5000 yrs ÷ 5730 yrs

    = 0.873

N = 150 g ÷ 0.5^0.873

   = 150 g × 0.546

   = 81.9 g

Therefore, the mass of C-14 left after 5000 yrs is 81.9 g

5 0
3 years ago
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