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3 years ago

7

An ionic bond would most likely form between atoms of which of these pairs? A. oxygen and oxygen B. magnesium and oxygen C. magn

esium and magnesium User: What is an ionic bond? A. the attraction positive ions have for free electrons B. the force that holds the valence electrons to the atom C. the attraction and bonding of ions with equal opposite charges D. the bonding of ions to the same element atoms of neutral charge

1 answer:

RoseWind [281]3 years ago

7 0

Qn1)B
qn2)C

This is because a nonmetal(oxygen) and a metal(magnesium) form ionic bond.
An ionic bond is strong electrostatic forces of attraction between positive and negative ions (opposite ions)

exp: Mg2+ + O2- ---> MgO

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There are 6 oxygen atoms in a half dozen water (h2o) molecules. how many hydrogen atoms are in the same half dozen molecules?

dlinn [17]

There are 12 hydrogen atoms in 6(H2O)

5 0

3 years ago

How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?

Rainbow [258]

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

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 = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C × 17.3°C

Q = 2640.96 J

5 0

2 years ago

Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2

lorasvet [3.4K]

Al

Explanation:

The limiting reactant will be Al:

4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

Given parameters:

Mass of Al = 30g Molar mass = 27g/mol

Number of moles = $\frac{mass}{molar mass}$ = $\frac{30}{27}$

Number of moles of Al = 1.111 mole

Mass of O₂ = 30g, molar mass = 32g/mol

Number of moles = $\frac{30}{32}$ = 0.94mol

In the reaction:

4 moles of Al reacted with 3 moles of O₂

1.11moles of Al will require $\frac{1.11 x3}{4}$ = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0

2 years ago

g In the absence of allosteric effectors, the enzyme phosphofructokinase displays Michaelis–Menten kinetics (see Fig. 7.15). The

Furkat [3]

Answer:

The value of the Michaelis–Menten constant is 0.0111 mM.

Explanation:

Michaelis–Menten 's equation:

$v_o=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}$

$V_{max}=k_{cat}[E_o]$

Where:

$v_o$ = rate of formation of products

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

On substituting all the given values

We have :

$\frac{v_o}{V_{max}}=0.90$

[S] = 0.10 mM

$\frac{v_o}{V_{max}}=\frac{[S]}{(K_m+[S])}$

$0.90=\frac{0.10 mM}{K_m+0.10 M}$

$K_m=0.0111 mM$

The value of the Michaelis–Menten constant is 0.0111 mM.

6 0

2 years ago

Iron-59 is a beta emitter with a half-life of 44.5 days. If a sample initially contains 180 mg of iron-59, how much iron-59 is l

Jobisdone [24]

Answer:

2.79 mg iron-59 is left in the sample after 267 days.

Explanation:

Given that:

Half life = 44.5 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{44.5}\ days^{-1}$

The rate constant, k = 0.0156 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration = 180 mg

Time = 267 days

So,

$[A_t]=180\ mg\times e^{-0.0156\times 267}$

$[A_t]=180\times e^{-0.0156\times 267}\ mg$

$[A_t]=2.79\ mg$

<u>2.79 mg iron-59 is left in the sample after 267 days.</u>

5 0

2 years ago