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3 years ago

#NEEDHELPASAP #MARINESCIENCE

Chemistry

1 answer:

Keith_Richards [23]3 years ago

3 0

I believe it is B because surfers could tell when, they seat on the board and feel the water

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Circle the correct one from given alternatives.

mash [69]

Answer:

24 is the correct anwer

this the anwer text this u no

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2 years ago

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How many formula units of CaO are in 32.7 g of CaO?

max2010maxim [7]

Answer:

3.51× 10²³ formula units

Explanation:

Given data:

Mass of CaO = 32.7 g

Number of formula units = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass/molar mass

Number of moles = 32.7 g/ 56.1 g/mol

Number of moles = 0.583 mol

Number of formula units:

1 mole = 6.022 × 10²³ formula units

0.583 mol × 6.022 × 10²³ formula units / 1 mol

3.51× 10²³ formula units

The number 6.022 × 10²³ is called Avogadro number.

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3 years ago

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

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3 years ago

1)¿Como estoy siendo responsable con el cuidado de mi salud y de los demás , ya sea en casa o si algo a realizar alguna activida

erica [24]

Answer:

The most effective strategy to control the spreading of some infections is through social distancing.

Explanation:

Since we don't have an effective treatment to stop the infection of some pathogens including the recent pandemic SARS-Cov 2 virus, it is imperative to prevent it from spreading

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3 years ago

Does the density of ice affect the melting rate of ice, or does adding objects affect melting rates? I'm going to need evidence

kotykmax [81]

The density of ice does not affect the melting rate. But, adding an object does affect the melt rate. The reason this is is because when there is an object, there is less to melt. Hence, affecting the melting rate.

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3 years ago

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