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4 years ago

5

What is the molecular formula for a compound with an empirical formula of CH2O if its experimentally determined molecular weight

is 180 amu?

1 answer:

NNADVOKAT [17]4 years ago

3 0

Answer:

C6H12O6.

Explanation:

Adding up the relative atomic masses of the elements

C + 2H + O

= 12 + 2*1 + 16.

= 30.

180 / 30 = 6 so the molecular formula is:

C6H12O6.

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Can torque rods desaturate reaction wheels

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3 years ago

Write the beta decay equation for the following isotope: 15/6 C?

vesna_86 [32]

Answer:

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{N}$

Explanation:

The unbalanced nuclear equation is

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

15 = 0 + A, so A = 15 - 0 = 15, and

6 = -1 + Z, so Z = 6 + 1 = 7

Then, your nuclear equation becomes

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{X}$

Element 7 is nitrogen, so the balanced nuclear equation is

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{N}$

8 0

4 years ago

Which of the following are likely to form a covakent bond

Dmitriy789 [7]

Answer:

For the most part, non-metals (excluding Nobel gases) are the most likely to form covalent bonds. Pure covalent bonds are formed between atoms with the same electronegativity, ie. they are trying to hold on to the electrons in the bond with the same strength.

5 0

4 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

<u>T₂ = 19.95°C</u>

6 0

3 years ago