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ser-zykov [4K]
3 years ago
6

Who was the old woman that was in the temple during the presentation of Jesus​

Chemistry
1 answer:
Lynna [10]3 years ago
5 0

Answer:Anna

Explanation:Anna the Prophetess is a woman mentioned in the Gospel of Luke. According to that Gospel, she was an elderly woman of the Tribe of Asher who prophesied about Jesus at the Temple of Jerusalem. She appears in Luke 2:36–38 during the presentation of Jesus at the Temple.

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2074 Set B Q.No. 1 What mass of nitrogen will be requires
timurjin [86]

140 g of nitrogen (N₂)

Explanation:

We have the following chemical equation:

N₂ + 3 H₂ -- > 2 NH₃

Now, to find the number of moles of ammonia we use the Avogadro's number:

if        1 mole of ammonia contains 6.022 × 10²³ molecules

then   X moles of ammonia contains 6.022 × 10²⁴ molecules

X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³

X = 10 moles of ammonia

Taking in account the chemical reaction we devise the following reasoning:

If        1 mole of nitrogen produces 2 moles of ammonia

then  Y moles of nitrogen produces 10 moles of ammonia

Y = (1 × 10) / 2

Y = 5 moles of nitrogen

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of nitrogen (N₂) = 5 × 28 = 140 g

Learn more about:

Avogadro's number

brainly.com/question/13772315

#learnwithBrainly

7 0
4 years ago
How is the periodic table structured with regard to elements with similar properties ?
sleet_krkn [62]
The families, or groups, have similar properties, for example, the first group 1 are made of extremely reactive metals. They will also have similar boiling, melting, and conductivity. With valence electrons all in the same orbital (alkaline metals all have valence electrons in the s orbital)<span />
6 0
3 years ago
Read 2 more answers
9. Gasoline burns easily if ignited
Elena L [17]

Answer:

I think it is chemical change

5 0
3 years ago
In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera
vredina [299]

Answer : The enthalpy change of dissolution of KClO_4 is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 1.55J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 107.70 g

\Delta T = change in temperature = T_2-T_1=(22.80-20.34)=2.46^oC

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

q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]

q=1112.3J=1.1123kJ

Now we have to calculate the enthalpy change of dissolution of KClO_4

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of KClO_4 = 2.84 g

Molar mass of KClO_4 = 138.55 g/mol

\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole

\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole

Therefore, the enthalpy change of dissolution of KClO_4 is 54.3 kJ/mole

5 0
4 years ago
What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?
zubka84 [21]
Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g 
H: 16.3% = 16.3 g 

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

C:  \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol

H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

C:  \dfrac{6.975}{6.975} = 1

H:  \dfrac{16.3}{6.975} \approx 2.3

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark

I hope this helps. =)
8 0
3 years ago
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