Q1. They are highly reactive. Q2. High reactivity, nonmetallic. Q3. Oxygen has an ion charge of -2. Q4. LiCl I believe. Q5. How electrons are shared. Q6 1. Q7. Share 2 valence electrons, I believe.
Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
<em>The volume of the glucose solution to be prepared</em> = 500 
<em>Molarity of the glucose solution to be prepared</em> = 1 M
i. Molar mass of glucose (
) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii.<em> mole = molarity x volume</em>. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole
BF3 .... BP = −100.3 °C
<span>RbCl ..... solid </span>
<span>CH3SCH3 ..... BP = 35-41 °C </span>
<span>SbH3 .... BP = −17 °C </span>
<span>SiS2 ..... solid </span>
<span>Ethanol solid --> ethanol melts --> ethanol liquid </span>
<span>-135C ---------------> -114C --------------> -50C </span>
<span>............ ΔT = 21C ....... ....... ΔT = 64C </span>
Answer:
please mark as brainliest!!
Explanation:
// C++ program to print initials of a name
#include <bits/stdc++.h>
using namespace std;
void printInitials(const string& name)
{
if (name.length() == 0)
return;
// Since touuper() returns int, we do typecasting
cout << (char)toupper(name[0]);
// Traverse rest of the string and print the
// characters after spaces.
for (int i = 1; i < name.length() - 1; i++)
if (name[i] == ' ')
cout << " " << (char)toupper(name[i + 1]);
}
// Driver code
int main()
{
string name = "prabhat kumar singh";
printInitials(name);
return 0;
}
Answer:
<h2>
<em><u>Data</u></em></h2>
Explanation:
Data are the facts, figures, and other evidence gathered through observations.
