All categories

2 years ago

HELP ME PLEASE

Chemistry

1 answer:

lions [1.4K]2 years ago

3 0

Answer:

Explanation:

Definitely not in the cell theory

You might be interested in

The state of matter represented on the right (letter D) has the

jek_recluse [69]

Answer:

A is the answer.

Explanation:

gorgeousness

5 0

2 years ago

WHAT IS THE COLOR OF METHYL ORANGE 1. YELLOW 2. PINK 3. ORANGE

Leya [2.2K]

Answer:

3.Orange it literally say's it methyl ORANGE

Explanation:

Brainliest plz

7 0

2 years ago

Read 2 more answers

The density of plutonium is 19.84 g/cm². What is its mass if the container has volume of 250 cm?

serious [3.7K]

Answer:

the answer is 7.936X 10⁻⁷kg.... the unit for volume must be cm³

Explanation:

we know,

ρ =m/v

given,

ρ =19.84/1000X(100)²= 1.984X10⁻⁶kg/m²

V (volume of the container means the volume of the element)=2.5 m³

NOW putting in the equation

1.984X10⁻⁶/2.5=m

or, m=7.936X 10⁻⁷kg(ANS)

thank u(i am not 100% sure but think it shoud be the answer)

4 0

2 years ago

Which element is the most reactive? A.sodium B.nickel C.carbon D.oxygen

ipn [44]

Answer: A. Sodium
Explanation: the most reactive element is sodium

7 0

3 years ago

An ice cube of mass 8.5g is added to a cup of coffee, whose temperature is 95 degrees Celcius and which contains 120 g of liquid

Ne4ueva [31]

Answer:

$\text { The temperature of the coffee after the ice melts is } 83.4^{\circ} \mathrm{C}$

Explanation:

Firstly determine heat absorbed by the ice-cube while it is melting:

$\mathrm{q}_{\mathrm{ice}}=\frac{8.5 \mathrm{gm} \text { ice } \times 1 \mathrm{mol} \text { ice } \times 6 \mathrm{kJ}}{18.0152 \mathrm{gm} \text { ice } \times 1 \mathrm{mol} \text { ice }}$

$\mathrm{q}_{\mathrm{ice}}=\frac{51}{18.0152}$

$\mathbf{q}_{\text {ice }}=2.83 \mathrm{kJ}$

$\text { Now heat emitted by the coffee while ice is melting: } q_{\varepsilon}=-q_{\mathrm{ice}}=-2.83 \mathrm{kJ}$

Now the temperature of the coffee when the ice has melted:

$\mathrm{q}_{\mathrm{c}}=\mathrm{m}(\Delta \mathrm{T})(\mathrm{C})$

$\Delta \mathrm{T}=\frac{q_{c}}{m c}$

$\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}=\frac{q_{c}}{m c}$

$\mathrm{T}_{\mathrm{f}}=\left(\frac{q_{\mathrm{c}}}{m c}\right)+\mathrm{T}_{\mathrm{i}}$

$\mathrm{T}_{\mathrm{f}}=\frac{-2.83 \times 10^{3} \mathrm{J}}{120 \mathrm{gm}\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}+95^{\circ} \mathrm{C}$

$T_{f}=\frac{-2.83 \times 10^{3}}{501.6}+95$

$\mathrm{T}_{\mathrm{f}}=-5.64+95$

$\mathrm{T}_{\mathrm{f}}=89.37^{\circ} \mathrm{C}$

Since both the substances eventually reach thermal equilibrium then they both must have the same final temperature.

$\text { As } \mathrm{q}_{\mathrm{w}}=-\mathrm{q}_{\mathrm{c}}$

By solving this, equation achieved is:

$\mathrm{T}_{\mathrm{f}}=\frac{\mathrm{m}_{\mathrm{w}} \mathrm{T}_{\mathrm{iv}} \mathrm{C}_{\mathrm{w}}+\mathrm{m}_{\mathrm{c}} \mathrm{T}_{\mathrm{ic}} \mathrm{C}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{W}}+\mathrm{m}_{\mathrm{c}} \mathrm{C}_{\mathrm{c}}}$

$\mathrm{T}_{\mathrm{f}}=\frac{(8.5 \mathrm{gm})\left(0.0^{\circ} \mathrm{C}\right)\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)+(120 \mathrm{gm})\left(89.37^{\circ} \mathrm{C}\right)\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}{(8.5 \mathrm{gm})\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)+(120 \mathrm{gm})\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}$

$\mathrm{T}_{\mathrm{f}}=\frac{0+44827.992}{35.53+501.6}$

$\mathrm{T}_{\mathrm{f}}=\frac{44827.992}{537.13}$

$\mathrm{T}_{\mathrm{f}}=83.4^{\circ} \mathrm{C}$

7 0

3 years ago