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4 years ago

5

What is the volume V of a sample of 2.30 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

is 8.92×103kg/m3.

1 answer:

djyliett [7]4 years ago

5 0

Answer:

1.636 x 10^-5 m^3

Explanation:

As we know that

1 mol of copper = 63.5 g of copper

2.3 mol of copper = 63.5 x 2.3 = 146.05 g of copper = 0.146 kg of copper

Density of copper = 8.92 x 10^3 kg/m^3

The relation between mass and volume is given by

mass = volume x density

$Volume = \frac{mass}{density}$

$Volume = \frac{0.146}{8.92\times10^3}=1.636\times10^{-5}m^3$

Thus, the volume of copper is 1.636 x 10^-5 m^3.

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Convert .00016 to scientific notation

Sholpan [36]

Answer:

$1.6 \,\,\,10^{-4}$

Explanation:

Recall that scientific notation involves expressing the number in powers of ten, and using a multiplicative factor whose value is larger than or equal to 1 (one) and strictly smaller than 10 in the form:

$a\,\,\,10^n\\with\,\,\,1\leq a$

and with n an integer value.

in our case, notice that 0.00016 = 16 / 100000 = 1.6 / 10000

and that denominator can be expressed as: 10^4, and can also be brought to the numerator as a negative power of ten:

1.6 / 10000 = 1.6 / 10^4 = 1.6 * 10^(-4)

$\frac{16}{100000} =\frac{1.6}{10000}=\frac{1.6}{10^4}=1.6\,\,\,10^{-4}$

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3 years ago

The cation that is reabsorbed from the urine in response to aldosterone

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Answer:

If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.

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3 years ago

The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus)

Artemon [7]

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge $q=5\sin4\pi t\ mC$

Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

We need to calculate the current

Using formula of current

$i(t)=\dfrac{dq}{dt}$

Put the value of charge

$i(t)=\dfrac{d}{dt}(5\sin4\pi t)$

$i(t)=5\times4\pi\cos4\pi t$

$i(t)=20\pi\cos4\pi t$

(a).We need to calculate the power delivered to the element

Using formula of power

$p(t)=v(t)\times i(t)$

Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

$p(t)=60\pi\times10^{-3}\cos^2(4\pi t)$

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

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kiruha [24]

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mars1129 [50]

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