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2 years ago

Is this equation balanced? 2HCl + CaCO3 → CaCl2 + H2O + CO2

Physics

1 answer:

Serggg [28]2 years ago

7 0

Answer:

Yes it is

Explanation:

the sum moles at the left side equals the sum of moles at the right side

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As 390 g of hot milk cools in a mug, it transfers 30,000 J of heat to the environment. What is the temperature change of the mil

kotykmax [81]

The correct answer is 19.72 °C. The first step is to determine the amount of heat that was lost per gram (30,000 J) / (390 g) = 76.92 J/g. Then to determine the temperature change, divide 76.92 J/g with the given specific heat of milk (76.92 J/g) / (3.9 J/g°C) = 19.72 °C.

5 0

3 years ago

Read 2 more answers

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

Radiant energy comes from the _______.

snow_lady [41]

Answer:

C. Sun

Explanation:

7 0

2 years ago

Read 2 more answers

If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

7 0

2 years ago

Write the ground-state electron configurations of the following ions. (a) Li+ (b) N3− (c) In3+ (Use the noble gas core electron

maria [59]

Li+  [He]

N³-  [Ne]

In³+  [Kr] 4d10

Tl+  [Xe] 4f14 5d10 6S2

6 0

3 years ago

Read 2 more answers