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3 years ago

I need it done assapp

Chemistry

2 answers:

Alja [10]3 years ago

4 0

Answer:

water molecules

Explanation:

Ivenika [448]3 years ago

3 0

Answer:

State: Liquid

Explanation:

In this state molecules move freely and slide past each other.

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Camels store the fat tristearin in the hump

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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

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3 years ago

Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change

USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation follows:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]$

We are given:

$\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]$

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

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3 years ago

What are the reactants and products? Write a reaction equation and label the reactants and products.

djyliett [7]

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