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3 years ago

FAN AND MEDAL

2 answers:

3 0

The correct answer is false. A compound microscope can magnify something up to 200 times and further. A compound microscope is a microscope with more than one lens and its own source of light. Today's strongest compound microscopes have magnifying powers of 1000 times to 2000 times.

Ulleksa [173]3 years ago

3 0

The correct answer is False

Explanation:

A microscope is an instrument used in many scientific fields including biology and chemistry to magnify the image of an object and therefore study and look at it, which is not possible without a microscope due to the limitations of the human eye. Additionally, a compound microscope is a type of microscope that includes multiple lenses to magnify objects and therefore, allows scientist to observe the same object with different sizes. About this, it is important to mention the range objects can be magnified depends on the type of microscopes but in general terms, this type of microscope include lenses that magnify objects from 10 times to around 400 times, and especial microscopes can magnify objects up to 1000 times. Thus, it is false a compound microscope can magnify something only up to 200 times.

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** worth 20 points + brainliest **

Bad White [126]

Answer : The correct option is, (D) $83^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat released = -1300 J

m = mass of water = 40 g

c = specific heat of water = $4.18J/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $91.0^oC$

Now put all the given values in the above formula, we get the final temperature of water.

$-1300J=40g\times 4.18J/g^oC\times (T_{final}-91.0^oC)$

$T_{final}=83.2^oC\approx 83^oC$

Therefore, the final temperature of the water is, $83^oC$

5 0

3 years ago

A balanced chemical equation is a direct presentation of the ___________.

Lisa [10]

Answer:

The chemical equation needs to be balanced so that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation:

How do you balance equations?

In general, however, you should follow these steps:

1. Count each type of atom in reactants and products.

2. Place coefficients, as needed, in front of the symbols or formulas to increase the number of atoms or molecules of the substances.

3. Repeat steps 1 and 2 until the equation is balanced.

5 0

3 years ago

How many grams of KBr are present in 250.0mL of a 1.20M solution

castortr0y [4]

Molar mass KBr = 119 g/mol

Volume in liters: 250.0mL / 1000 => 0.25 L

n = M x V

n = 1.20 x 0,25 => 0.3 moles of KBr

Therefore:

1 mole KBr ----------- 119 g
0.3 moles KBr -------- ??

0.3 x 119 / 1 => 35.7 g of KBr

4 0

3 years ago

How many orbitals does the s subshell contain? <br> 1<br> 3<br> 5<br> 7

Kryger [21]

Answer:1

Explanation: There is only one orbital in the s sublevel. It can hold up to two electrons.

6 0

3 years ago

Please answer, this is due in 30 minutes

notsponge [240]

Answer:

0.591 g of magnesium phosphate is the theoretical yield.

Magnesium nitrate is the limiting reactant.

Explanation:

Hello!

In this case, since the balanced reaction turns out:

$3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3$

Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:

$m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2$

Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.

However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:

$Y=\frac{actual}{0.591g}*100\%$

Best regards!

4 0

3 years ago