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makvit [3.9K]
3 years ago
14

When working with hazardous substances in the laboratory your partner splashes a small amount of Sodium Hydroxide onto their clo

thing, you immediately help them to the safety shower to rinse off for 15 minutes
a.) After you help your partner into the safety shower, what should be your next step?



b.) What should you have with you when reporting the incident to emergency responders?
Chemistry
1 answer:
alexira [117]3 years ago
5 0

Answer:

a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.

b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.

You might be interested in
Please help!!!!
stealth61 [152]

Answer:

Cells are extremely small.

Explanation:

As Mendel describes in this story, cells are so small they cannot normally be seen with the naked eye. ... The total organism remains the same throughout this process, and (usually) has a longer time on earth than any one of its cells.

3 0
3 years ago
A 21.0 kg child is running at a velocity of 2.50 m/s. The child is carrying a 2.25 kg gallon of milk. What is the momentum of th
KIM [24]

Answer: The momentum of the child and milk together is 58.125 kg.m/s

Explanation:

Momentum is defined as the product of object's mass and velocity.

Mathematically,

p=mv

where, p = momentum

m = mass of the object

v = velocity of the object

In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:

p=(m_{child}+m_{milk})v     ....(1)

Given:

m_{child}=21.0kg\\m_{milk}=2.25kg\\v=2.5m/s\\p=?kg.m/s

Putting values in equation 1, we get:

p=(21+2.25)kg\times 2.5m/s=58.125kg.m/s

Hence, the momentum of the child and milk together is 58.125 kg.m/s

7 0
4 years ago
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
The density of a metal cube is 8 x103 kg/m3. Convert
Rom4ik [11]

Answer:

Explanation:

[peirod4

4 0
3 years ago
A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X ​
Tasya [4]

\LARGE{ \boxed{  \rm{ \red{Required \: answer}}}}

☃️ Chemical formulae ➝ \sf{CaCO_3}

<h3><u>How to find?</u></h3>

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

\boxed{ \sf{No. \: of \: moles =  \frac{given \: weight}{molecular \: weight} }}

<h3><u>Solution:</u></h3>

Atomic weight of elements:

Ca = 40

C = 12

O = 16

❍ Molecular weight of \sf{CaCO_3}

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

3 0
4 years ago
Read 2 more answers
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