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3 years ago

What are things that cause change

Chemistry

2 answers:

inessss [21]3 years ago

5 0

Explanation:

A catalyst means "something that causes activity,an event, or change." And usually,these events and changes are big.

I hope it's helpful!

shutvik [7]3 years ago

5 0

Ummmmmm what ? It’s kinda confusing

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Explain what needs to happen for the box to start moving

VMariaS [17]

Answer:

It needs an force in the opposite direction.

Explanation:

Example:

The box is on a flat surface. You use your hands to push it. You have to push it in the opposite direction.

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3 years ago

How much does a gallon of water WEIGH?

elixir [45]

A gallon of water weighs 8.34 pounds

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4 years ago

Arrange follwoing substances from lowest to highest lattice energy" MgS, KI, GAN, LiBr

sleet_krkn [62]

Answer: KI, LiBr, MgS,GaN

Explanation:

Ionic sizes are the deciding factor in lattice energy. The smaller the ion, the higher the lattice energy. Since K+> Li+ and I- is larger than Br-, LiBr will have a higher lattice energy. The size of the sulphide ion is much smaller than the nitride ion but the Gallium is much smaller due to lanthanide contraction hence the answer.

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3 years ago

Which state is true about the speed of heat transfer by conduction?

Thepotemich [5.8K]

I think the answer is, Thermal? I hope u get it right

3 0

3 years ago

A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con

Nataly_w [17]

Answer:

$\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

$Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}$

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

$Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J$

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

$n=1.400g*\frac{1mol}{89.05g} =0.01572mol$

Thus, we obtain:

$\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Best regards!

7 0

3 years ago