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3 years ago

7

What is the name of the compound that goes with the chemical formula MgCl2

1 answer:

frez [133]3 years ago

8 0

Magnesium chloride is your answer

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We decided to make an iced latte by adding ice to a 200 mL hot latte at 45 °C. The ice starts out at 0 C. How much ice do we nee

tankabanditka [31]

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = $m s\Delta T$

$Q_1 = (200)(4.186)(45 - 10)$

$Q_1 = 29302 J$

now heat absorbed by ice is given as

$Q_2 = mL + ms\Delta T$

$Q_2 = m(335 + 4.186(10 - 0))$

$Q_2 = m(376.86)$

now by heat balance we have

$Q_1 = Q_2$

$29302 = m(376.86)$

$m = 77.75 g$

6 0

3 years ago

In a closed energy system, energy is?

Alisiya [41]

Answer:

Explanation:

A closed system can exchange energy but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.

Hope this helped you!

3 0

3 years ago

Q1. Which statement is correct

marishachu [46]

Answer:

C

D

Explanation:

7 0

3 years ago

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las

topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

$w=\frac{2.44 \lambda L}{D}$

where $\lambda$ is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

$L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m$

4 0

3 years ago

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

4 0

3 years ago