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2 years ago

Disease caused by virus are more dangerous.Why?

Chemistry

1 answer:

nlexa [21]2 years ago

7 0

Answer:

Explanation:

1. Virus's are hard to detect because of their simple construction.

2. Some mutate very easily.

3. It is hard to isolate the virus and kill it without doing damage to the host.

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How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

1gallon=3.79 liters, The gas tank of a car holds 15 gallons.if you were traveling in Europe how many liters if petrol would you

Andrei [34K]

For an approximate result, multiply the volume value by 3.785
Answer ≈ 56.7812

6 0

3 years ago

Read 2 more answers

Which formula can be used to calculate the theoretical yield?

Bas_tet [7]

Answer:

Option C, (Actual yield ÷ percent yield) × 100

Explanation:

Theoretical yield is defined as the total amount of product formed for given reactants in a chemical reaction. It is an ideal case which assumes no exceptions or wastage.

The mathematical relation between the actual yield, percent yield and theoretical yield is as follows -

$P.Y. = \frac{M_{A.Y.}}{M_{T.Y.}} * 100$

Where

P.Y. represents the percent yield a

M A.Y. represents the mass obtained from actual yield

M T.Y. represents the mass obtained from theoretical yield

Hence, if we rearrange the formula, we get -

$M_{T.Y.} = \frac{M_{A.Y.}}{P.Y.} * 100$

Hence, option C is correct

5 0

2 years ago

A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?

kotykmax [81]

First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass

mass = 252.4795g however the 6.5L has only 2 sig figs so i would say

mass CaCl2 = 2.5 * 10 ^2 g

5 0

3 years ago

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What is the primary requirement for a molecule to be analyzed by Gas Chromatography?

BabaBlast [244]

Answer:

b) The molecule has a molecular weight under 200 g/mole

Explanation:

The molecule has a molecular weight under 200 g/mole is the primary requirement for a molecule to be analyzed by Gas Chromatography.

3 0

3 years ago

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