All categories

3 years ago

Name the following ionic compounds: BeCl2

Physics

1 answer:

tatyana61 [14]3 years ago

4 0

Answer:

Beryllium chloride

Explanation:

I hope this helps!

Can I get branliest?!

You might be interested in

An object is moving on a horizontal frictionless surface. if the net force applied to the object in the direction of motion is d

sukhopar [10]

Answer:

doubled

Explanation:

F=ma1----------(1)

2F = ma2-------(2)

Divide 2nd equation by 1st one

we get a1×2=a2

5 0

2 years ago

A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work i

podryga [215]

The answer is 3) 480 joules

6 0

3 years ago

A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is

Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10² N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x) = 0.5 × k × x² ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x) = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0

3 years ago

When an 8 V battery is connected to a resistor, a 2 A current flows in the resistor. What is the resistor's value?

Vinvika [58]

Answer:

Explanation:

V=IR I= curren V=volt R=resistor

8=2.R 8/2=R R=4

5 0

3 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

4 0

3 years ago

Name the following ionic compounds: BeCl2​

Name the following ionic compounds: BeCl2